All Solutions
Page 876: Assessment
b) The best explanation is B and it is supported by Wien’s law and the relation between frequency and wavelength.
$$
T=frac{nu_{peak}}{5.88times 10^{10}text{ Hz/K}}.
$$
b) Since the greater the wavelength the lower the frequency this order is the inversion of that in part a) so we have: red, green, blue.
c) We know that not all of those photons are equally energetic due to different frequency. Since we need more of the less energetic photons emitted per second to get the same power this order is the inversion of the order of increasing energy (and thus the inversion of that of increasing frequency in part b)) so it reads: blue, green, red.
b) The best explanation is B. since the energy of a single photon depends only on its’ frequency.
b) The best explanation is A. due to the property that “yellow” photon’s are less energetic than “blue” ones.
$f_{peak} = 1.82 times 10^{14}$ Hz
$textbf{Find:}$
Surface temperature of Betelgeuse, $T$
To calculate $T$, we use the Wien’s displacement law as follows:
$$
begin{align*}
f_{peak} &= 5.88 times 10^{10} text{ Hz/K} times T \
T &= dfrac{f_{peak}}{5.88 times 10^{10}} \
&= dfrac{1.82 times 10^{14}}{5.88 times 10^{10}} \
&=3095 K
end{align*}
$$
Hence, the surface temperature of Betelgeuse is $boxed{text{3095 K}}$
T = text{3095 K}
$$
$$
nu_{peak} = 5.88times 10^{10}text{ Hz/K}times T
$$
giving for the temperature of the star
$$
T = frac{nu_{peak}}{ 5.88times 10^{10}text{ Hz/K}} = frac{1.82times10^{14}text{ Hz}}{ 5.88times 10^{10}text{ Hz/K}} = 3095text{ K}.
$$
$T = 95^ circ text{C} = 308 text{ K}$
$textbf{Find:}$ Wavelength of radiation, $lambda$
$$
textbf{Formula:}
$$
$$
begin{gather}
f_{peak} = 5.88 times 10^{10} text{ Hz/K} times T \
lambda = dfrac{c}{f}
end{gather}
$$
We first calculate for $f_{peak}$ using Eq(1) as follows:
$$
begin{align*}
f_{peak} &= 5.88 times 10^{10} times 308 text{ K} \
&= 1.81 times 10^{13} text{ Hz} \
text{To calculate for } lambda text{ we use Eq(2):} \
lambda &= dfrac{c}{1.81 times 10^{13}} \
&= dfrac{3.0 times 10^8}{1.81 times 10^{13}} \
&= boxed{1.66 times 10^{-5} text{ m}}
end{align*}
$$
$$
T = 95^circ text{ F} = (95+460)timesfrac{5}{9}text{ K}=308text{ K}.
$$
Using Wien’s law we get
$$
nu_{peak} = (5.88times10^{10}text{ Hz/K})times 308text{ K} = 1.81times 10^{13}text{ Hz}.
$$
The relation between frequency and wavelength is
$$
lambda_{peak} = frac{c}{nu} =16.6text{ $mu$m}.
$$
$T = 2.7 K$
a) To find the peak frequency, we use Wien’s displacement law as follows:
$$
begin{align*}
f_{peak} &= 5.88 times 10^{10} text{ Hz/K} times T \
&= 5.88 times 10^{10} text{ Hz/K} times 2.7 text{ K} \
&= boxed{1.59 times 10^{11} text{ Hz}}
end{align*}
$$
$$
begin{align*}
lambda &= dfrac{c}{f_{peak}} \
&= dfrac{3.0 times 10^{8}}{1.59 times 10^{11}} \
&= boxed{1.88 times 10^{-3} text{ m}}
end{align*}
$$
b) $lambda =1.88 times 10^{-3} text{ m}$
$$
nu_{peak}=5.88times10^{10}text{ Hz/K}times T = 1.6times 10^{11}text{ Hz}.
$$
b) From the relation between wavelength and frequency we obtain
$$
lambda_{peak}=frac{c}{nu_{peak}}=frac{3times10^8text{ m/s}}{1.6times 10^{11}text{ Hz}}=1.9text{ mm}.
$$
$E = 6.5 times 10^{-19}$ J
$textbf{Find:}$ frequency, $f$ and wavelength, $lambda$
$textbf{Formula:}$
$$
begin{gather}
f = dfrac{E}{h} \
lambda = dfrac{c}{f}
end{gather}
$$
Using Eq(1), we can calculate for $f$ as follows:
$$
begin{align*}
f &= dfrac{6.5 times 10^{-19}}{6.626 times 10^{-34}} \
&= boxed{9.8 times 10^{14} text{ Hz}}
end{align*}
$$
Using Eq(2), we can calculate for $lambda$ as follows:
$$
begin{align*}
lambda &= dfrac{3.0 times 10^8}{9.8 times 10^{14}} \
&= boxed{3.06 times 10^{-7} text{ m}}
end{align*}
$$
$$
lambda = 3.06 times 10^{-7} text{ m}
$$
$P = 2.5$ W
$f = 5.2 times 10^{14}$ Hz
$textbf{Find}$ the number of photons in a second, $n$
$textbf{Formula:}$
$$
begin{gather}
P = dfrac{E}{t} \
E = nhf
end{gather}
$$
To obtain energy, $E$ we use Eq(1) such that:
$$
begin{align*}
E &= P times t \
&= 2.5 text{ W} times 1 text{ s} \
&= 2.5 text{ J}
end{align*}
$$
Then, using Eq (2), we can obtain $n$ as follows:
$$
begin{align*}
E &= nhf \
n &= dfrac{E}{hf} \
&= dfrac{2.5}{(6.626 times 10^{-34}) times 5.2 times 10^{14}} \
&= boxed{7.26 times 10^{18} text{ photons}}
end{align*}
$$
n = 7.26 times 10^{18} text{photons}
$$
$$
E_0 = hnu
$$
we can write
$$
2.5text{ J} = nhnu
$$
giving
$$
n=frac{2.5text{ J}}{hnu} =frac{2.5text{ J}}{6.62times10^{-34}text{ J s}times5.2times10^{14}text{ Hz}} = 7.26times 10^{18},
$$
so this source emits $7.26times 10^{18}$ photons per second.
$f = 9.95 times 10^{14}$ Hz
$KE_{max} = 0.180 times 10^{-19}$ J
$textbf{Find}$ work function, $W_o$
We calculate for $W_o$ as follows:
$$
begin{align*}
KE_{max} &= hf – W_o \
W_o &= hf – KE_{max} \
&= (6.626 times 10^{-34}) cdot (9.95 times 10^{14}) – (0.180 times 10^{-19}) \
&= boxed{6.41 times 10^{-19} text{ J}}
end{align*}
$$
W_o = 6.41 times 10^{-19} text{ J}
$$
$$
hnu=E_k + W_0.
$$
From here we obtain
$$
W_0 = hnu-E_k = 6.62times10^{-34}text{ J s}times 9.95times 10^{14}text{ Hz} – 0.180times 10^{ -19}text{ J} =6.41times 10^{ -19}text{ J}.
$$
$$
2.5text{ J} = nfrac{hc}{lambda}
$$
which yields
$$
n=frac{2.5text{ J}timeslambda}{hc} = frac{2.5text{ J}times 380times 10^{-9}text{ m}}{6.62times10^{ -34}text{ J}times3times10^8text{ m/s}} = 4.8times10^{18}.
$$
b) Again we have by substituting now the wavelength of red photons
$$
n=frac{2.5text{ J}timeslambda}{hc} = frac{2.5text{ J}times 750times 10^{-9}text{ m}}{6.62times10^{ -34}text{ J}times3times10^8text{ m/s}}=9.4times10^{18}.
$$
Einstein’s equation for the photoelectric effect is given by
$$
begin{align}
hnu=K+W
end{align}
$$
where $nu$ is the frequency of the photon, $K$ is the kinetic energy of the ejected electron, $W$ is the work function and $h=6.62times 10^{-34}text{Js}$ is Planck’s constant.
The relationship between the frequency of the photon $nu$ and its wavelength $lambda$ is given by
$$
begin{align}
nu=frac{c}{lambda}
end{align}
$$
where $c=3times 10^{8}frac{text{m}}{text{s}}$ is the speed of light in vacuum
The wavelength of the electromagnetic wave $lambda$ is
$$
begin{align*}
lambda=264text{nm}=264times 10^{-9}text{m}
end{align*}
$$
In this problem, it is required of us to calculate the work function $W$. Since the work function $W$ depends on the material (metal) and it does not depend on the kinetic energy of the electron $K$, we can put $K=0$ in Eq. (1). Then we have
$$
begin{align*}
hnu=W
end{align*}
$$
If we plug in Eq. (2) in the equation above, and plug in the numerical value for the wavelength of the wave $lambda$, we get
$$
begin{align*}
W&=frac{hc}{lambda}\
&=frac{6.62times 10^{-34}text{Js} times 3 times 10^8 frac{text{m}}{text{s}}}{264times 10^{-9}text{m}}\
&=7.5times 10^{-19}text{J}
end{align*}
$$
We can express this result in the units of electron-volts if we use the relationship
$$
begin{align*}
1text{J}=6.24 times 10^{18}text{eV}
end{align*}
$$
We get
$$
begin{align*}
boxed{W}&=7.5times 10^{-19}text{J}\
&=7.5times 10^{-19}times 6.24 times 10^{18}text{eV}\
&=46.8times 10^{-1}text{eV}\
&=boxed{4.68 text{eV}}
end{align*}
$$
W=4.68 text{eV}
$$
b) According to Wien’s law we can write
$$
nu_A = 5.88times10^{10}text{ Hz/K}times T_A;quadnu_B = 5.88times10^{10}text{ Hz/K}times T_B.
$$
Dividing these equations we get
$$
frac{nu_A}{nu_B} = frac{ 5.88times10^{10}text{ Hz/K}times T_A}{ 5.88times10^{10}text{ Hz/K}times T_B} = frac{T_A}{T_B} =0.36.
$$
$T_{standard} = 2900$ K
$T_{halogen} = 3400$ K
$textbf{Formula: }$
$$
begin{gather}
f_{peak} = 5.88 times 10^{10} text{ Hz/K} times T
end{gather}
$$
a) We calculate the $f_{peak}$ for the standard at halogen lightbulbs as follows:
$$
begin{align*}
f_{standard} &= (5.88 times 10^{10} text{ Hz/K}) times (2900 text{ K}) \
&= 1.70 times 10^{14} text{ Hz} \
f_{halogen} &= (5.88 times 10^{10} text{ Hz/K}) times (3400 text{ K}) \
&= 2.0 times 10^{14} text{ Hz}
end{align*}
$$
Since $f_{halogen} > f_{standard}$, the $textbf{halogen bulb has higher frequency}$
b) The halogen bulb’s frequency is closer to the frequency $5.5 times 10^{14} text{ Hz}$
b) The halogen bulb’s frequency is closer to the frequency $5.5 times 10^{14} text{ Hz}$
b) We will calculate peak frequencies of both lightbulbs
$$
nu_{standard} =5.88times10^{10}text{ Hz/K}times T_{standard}=1.7times10^{14};quad nu_{halogen} =5.88times10^{10}text{ Hz/K}times T_{halogen}=2times10^{14},
$$
so we see that halogen light bulb has peak frequency that is closer to what human eye perceives the best.
$$
nu_{peak} = 5.88times10^{10}text{ Hz/K}times 2850text{ K} =1.68times10^{14}text{ Hz} .
$$
b) We determine peak wavelength as
$$
lambda_{peak} = frac{c}{nu_{peak}} = frac{3times10^{8}text{ m/s}}{1.68times10^{14}text{ Hz}} = 1786text{ nm}
$$
which is deeply in infrared part of the spectrum so we expect much more infrared than ultraviolet radiation.
$$
omega=sqrt{frac{k}{m}}.
$$
Since the linear frequency is connected to the angular frequency by the relation $omega=2pinu$ we get
$$
nu=frac{omega}{2pi}=frac{1}{2pi}sqrt{frac{k}{m}}=frac{1}{2pi}sqrt{frac{1215text{ N/m}}{1.340times10^{-26}text{ kg}}} = 4.8times10^{13}text{ Hz}.
$$
b) The energy quantum is $hnu$ so we will have to add exactly this much energy to excite the molecule:
$$
Delta E=hnu=6.62times10^{ -34}text{ J s }times 4.8times10^{13}text{ Hz} = 3.17times10^{-20}text{ J}.
$$
b) Blue light has higher frequency and thus a photon of blue light has higher energy.
c) Red light bulb has to radiate $150text{ J}$ in one second and this energy is comprised of photons each having the energy of $E_0=hc/lambda_{red}$ so we can write
$$
150text{ J} = nfrac{hc}{lambda_{red}}
$$
yielding
$$
n=frac{150text{ J}timeslambda_{red}}{hc} = frac{150text{ J}times650times10^{-9}text{ m}}{6.62times10^{-34}text{ J s}times3times10^8text{ m/s}} =4.91times10^{20}.
$$
Blue light bulb has to radiate $25text{ J}$ in one second and this energy is comprised of photons each having the energy of $E_0=hc/lambda_{blue}$ so we can write
$$
25text{ J} = nfrac{hc}{lambda_{blue}}
$$
yielding
$$
n=frac{25text{ J}timeslambda_{blue}}{hc} = frac{25text{ J}times460times10^{-9}text{ m}}{6.62times10^{-34}text{ J s}times3times10^8text{ m/s}} =5.79times10^{19}.
$$
b) In general for a metal by putting zero kinetic energy into Einstein’s formula for photoelectric effect we get
$$
hnu_{cutoff} = W_0Rightarrow nu_{cutoff}=frac{W_0}{h}.
$$
Now substituting data for each metal from the problem we get
$$
nu_{Al} = frac{W_{Al}}{h} = frac{4.28times1.6times10^{-19}text{ J}}{6.62times10^{-34}text{ J s}} =1.03times10^{15}text{ Hz};
$$
$$
nu_{Ca}= frac{W_{Ca}}{h} = frac{2.87times1.6times10^{-19}text{ J}}{6.62times10^{-34}text{ J s}} =6.94times10^{14}text{ Hz}.
$$
$$
frac{hc}{lambda} = E_k+W_0Rightarrow E_k=frac{hc}{lambda}-W_0
$$
from where we see that for fixed $W_0$ the greater the wavelength the lower the kinetic energy (since we divide by lambda in the first term on the right) so we can say that beam B produces photoelectrons of greater frequency.
b) Using the formula we derived in part a) and substituting data given in the problem we get
$$
E_{kA} = frac{6.62times10^{-34}text{ J s}times 3times10^8text{ m/s}}{620times10^{-9}text{ m}}-1.9times1.6times10^{-19}text{ J} = 1.6times10^{-20}text{ J};
$$
$$
E_{kB} = frac{6.62times10^{-34}text{ J s}times 3times10^8text{ m/s}}{410times10^{-9}text{ m}}-1.9times1.6times10^{-19}text{ J} = 1.8times10^{-20}text{ J}
$$
$$
E_0=W_0+E_kRightarrow E_k=E_0-W_0.
$$
Since Cadmium has lower work function we subtract less from $E_0$ (see previous formula) so emitted electrons from Cadmium surface have greater kinetic energy.
b) We already derived
$$
E_k = E_0 – W_0 =frac{hc}{lambda}-W_0.
$$
Substituting values for both metals from the problem we have
$$
E_{kZn} = frac{6.62times10^{-34}text{ J s}times3times10^8text{ m/s}}{275times10^{-9}text{ m}} -4.33times1.6times10^{-19}text{ J} = 2.93times10^{-20}text{ J};
$$
$$
E_{kCd} = frac{6.62times10^{-34}text{ J s}times3times10^8text{ m/s}}{275times10^{-9}text{ m}} -4.22times1.6times10^{-19}text{ J} = 4.70times10^{-20}text{ J}.
$$
$$
hnu = E_k+W_0
$$
yielding
$$
E_k = hnu-W_0=6.62times10^{-34}text{ J s}times7.90times10^{14}text{ Hz}-2.24times1.6times10^{-19}text{ J} =1.65times10^{-19}text{ J}.
$$
b) To determine this we fill find the cutoff frequency by putting $E_k=0$ in Einsteins formula which gives
$$
hnu_{cutoff} = W_0Rightarrownu_{cutoff} = frac{W_0}{h} =frac{2.24times1.6times10^{-19}text{ J}}{6.62times10^{-34}text{ J s}} =5.41times10^{14}text{ Hz}
$$
and no lower frequency can eject electrons so the required range is
$$
(4.00times10^{14}text{ Hz},5.41times10^{14}text{ Hz}).
$$
$$
P=IA = Ileft(frac{d}{2}right)^2pi= 5.0times10^{-13}text{ W/m}^2times (4.25times10^{-3}text{ m})^2pi =2.84times10^{-17}text{ W}.
$$
This is the energy of $E=2.84times10^{-17}text{ J}$ comprised of photon passing each second. We can now write
$$
E=nhnu
$$
giving for the number of photons passing in one second
$$
n=frac{E}{hnu} =frac{2.84times10^{-17}text{ J}}{6.62times10^{-34}text{ J s}times7times10^{14}text{ Hz}}approx 60.
$$
(b) As car is accelerated, the speed of the car increases and consequently the momentum of the car increases. So the de Broglie wavelength will decrease, because the momentum is inversely proportional to the wavelength.
$$
begin{align}
lambda=frac{h}{p}
end{align}
$$
where $p$ is the momentum and $h$ is Planck’s constant.
a) If we change the momentum from $p$ to $2p$, ie. we double it, de Broglie’s wavelength changes in the following way
$$
begin{align*}
lambda rightarrow lambda’&=frac{h}{2p}\
&= frac{lambda}{2}
end{align*}
$$
The wavelength changes by a factor of $frac{1}{2}$ if we double the momentum $p$.
$$
begin{align*}
K=frac{mv^2}{2}
end{align*}
$$
We can write this equation in terms of the momentum $p=mv$, where $m$ is the mass of the particle:
$$
begin{align*}
K&=frac{mv^2}{2}\
&=frac{mp^2}{2m^2}\
&=frac{p^2}{2m}
end{align*}
$$
In order to relate de Broglie’s wavelength to the kinetic energy of the particle $K$, we express the momentum of the particle $p$ in terms of its kinetic energy $K$ from the equation above
$$
begin{align*}
p=sqrt{2mK}
end{align*}
$$
If we plug in the obtained result for $p$ in Eq. (1), we get
$$
begin{align*}
lambda&=frac{h}{p}\
&=frac{h}{sqrt{2mK}}
end{align*}
$$
If we doube the kinetic energy $K$, de Broglie’s wavelength changes in the following way:
$$
begin{align*}
lambda=frac{h}{sqrt{2mK}} rightarrow lambda’&=frac{h}{sqrt{4mK}}\
&=frac{lambda}{sqrt{2}}
end{align*}
$$
The wavelength changes by a factor of $frac{1}{sqrt{2}}$ if we double the kinetic energy of the particle $K$.
b) De Broglie’s wavelength $lambda$ changes by a factor of $frac{1}{sqrt{2}}$ if we double the kinetic energy of the particle $K$.
Smaller the wavelength greater the resolving power.
So electron microscope can resolve smaller object than a light microscope.
$lambda = dfrac{h}{mv} = dfrac{h}{p}$
A proton and an electron can have same de Broglie wavelength if their momentum is same.
Proton is 2000 times more massive than electron. So if the speed of the electron is 2000 times greater than that of proton, they will have same momentum and hence same de Broglie wavelength.
$lambda = dfrac{h}{mv}$ where $h$ is Planck’s constant.
The speed of the particle can be written in terms of the de Broglie wavelength as
$v = dfrac{h}{m lambda}$
The mass of the particle is $m = 6.69 times 10^{-27}:kg$.
The de Broglie wavelength of the particle is $lambda = 7.22:pm = 7.22 times 10^{-12}:m$.
The value of the Planck’s constant is $h = 6.626 times 10^{-34}:J:s$.
So the speed of the particle is given by
$v = dfrac{6.626 times 10^{-34}}{6.69 times 10^{-27} times 7.22 times 10^{-12}} = dfrac{6.626}{6.69 times 7.22} times 10^{-34} times 10^{27} times 10^{12}$
$$
v = 0.137 times 10^{5} = 1.37 times 10^4 :m/s
$$
v=1.37 times 10^4:m/s
$$
$lambda = dfrac{h}{mv}$
where $h (=6.626 times 10^{-34}:J:s$) is Planck’s constant.
So the velocity of the particle can be written as
$v=dfrac{h}{m lambda}$
Mass of the neutron is $m = 1.675 times 10^{-27}:kg$.
The de Broglie wavelength of the neutron is given as $lambda = 0.282:nm = 0.282 times 10^{-9}:m$.
The speed of the neutron is given by
$v = dfrac{6.626 times 10^{-34}}{1.675 times 10^{-27} times 0.282 times 10^{-9}} = dfrac{6.626}{1.675 times 0.282} times 10^{-34} times 10^{27} times 10^9$
$$
v = 14.03times 10^2:m/s = 1403 :m/s
$$
$lambda = dfrac{h}{mv} = dfrac{6.626 times 10^{-34}:J:s}{79 :kgtimes 4.2:m/s} = 0.02 times 10^{-34};m$
where $h = 6.626 times 10^{-34}:J.s$ is Planck’s constant.
lambda = 0.02 times 10^{-34}:m
$$
$lambda = dfrac{h}{mv} = dfrac{h}{p}$
So the momentum of the particle can be written as
$p = dfrac{h}{lambda}$.
The de Broglie wavelength of the electron is given as $lambda =0.15:nm$.
So its momentum is
$p = dfrac{6.626 times 10^{-34}}{0.15 times 10^{-9}} = dfrac{6.626}{0.15} times 10^{-34} times 10^9 =44.17 times 10^{-25}:kg:m/s$.
The kinetic energy of the electron is given by
$KE = dfrac{p^2}{2m_e}$
Mass of the electron is $m_e = 9.11 times 10^{-31}:kg$.
So, $KE = dfrac{(44.17 times 10^{-25})^2}{2 times 9.11 times 10^{-31}} = dfrac{(44.17)^2}{18.22} times 10^{-50} times 10^{31} =107.1 times 10^{-19}:J$
107.1 times 10^{-19}:J
$$
$lambda = 0.250:nm = 0.250 times 10^{-9}:m$.
So its momentum is given by
$p = dfrac{h}{lambda} = dfrac{6.626 times 10^{-34}}{0.250 times 10^{-9}} = dfrac{6.626}{0.250} times 10^{-34} times 10^{9} = 26.504 times 10^{-25} :kg:m/s$
Mass of the neutron is $m = 1.675 times 10^{-27}:kg$
So the speed of the neutron is given by
$v = dfrac{p}{m} = dfrac{26.504 times 10^{-25}}{1.675 times 10^{-27}} = dfrac{26.504}{1.675} times 10^{-25} times 10^{27} = 1582.3 :m/s$
$m_{electron} = 9.1 times 10^{-31}$ kg
$m_{proton} = 1.67 times 10^{-27}$ kg
The de Broglie wavelength can be calculated using the formula: $lambda = dfrac{h}{mv}$
Since the electron and proton have the same speed, their respective de Broglie wavelengths will depend on their masses. The proton is more massive than the electron, hence it has a shorter de Broglie wavelength.
$$
begin{align*}
ratio &= dfrac{m_{proton}}{m_{electron}} \
&= 1835
end{align*}
$$
$m_{proton} = 1835 m_{electron}$
As mentioned in a), the de Broglie wavelengths of the particles depend on their masses, hence:
$$
begin{align*}
dfrac{lambda_{electron}}{lambda_{proton}} &= dfrac{dfrac{h}{m_{electron}v}}{dfrac{h}{1835m_{electron}v}} \
&= boxed{1835}
end{align*}
$$
b) $dfrac{lambda_{electron}}{lambda_{proton}} = 1835$
De Broglie’s wavelength $lambda$ is given by
$$
begin{align}
lambda=frac{h}{p}
end{align}
$$
where $h$ is Planck’s constant and $p$ is the momentum.
Kinetic energy $K$ is given by, in terms of the momentum of the particle $p$
$$
begin{align}
K=frac{p^2}{2m}
end{align}
$$
where $m$ is the mass of the particle.
a) We will denote de Broglie’s wavelengths of the proton and the electron by $lambda_p$ and $lambda_e$, respectively. From the condition of the problem, we have
$$
begin{align*}
lambda_p=lambda_e=lambda
end{align*}
$$
In order to see which particle has greater kinetic energy, we have to express the momentum $p$ in terms of the kinetic energy $K$ and then plug in the obtained expression in Eq. (1). From Eq. (2), we see that the momentums of the proton $p_p$ and the electron $p_e$ are given by
$$
begin{align*}
p_p=sqrt{2m_p K_p}\
p_e=sqrt{2m_e K_e}
end{align*}
$$
where $m_p$ is the mass of the proton and $m_e$ is the mass of the electron. By plugging in these expressions into the condition $lambda_p=lambda_e$, we get
$$
begin{align*}
frac{h}{sqrt{2m_p K_p}}=frac{h}{sqrt{2m_e K_e}}
end{align*}
$$
Since the mass of the proton is greater than the mass of the electron $m_p>m_e$, we conclude that the kinetic energy of the electron must be greater than the kinetic energy of the proton $boxed{K_e>K_p}$.
$$
begin{align*}
K_p=frac{p_p^2}{2m_p}\
K_e=frac{p_e^2}{2m_e}
end{align*}
$$
If we use Eq. (1), we get (we also denote $lambda_p=lambda_e=lambda$, as in part a))
$$
begin{align*}
K_p=frac{h^2}{2m_p lambda^2}\
K_e=frac{h^2}{2m_e lambda^2}
end{align*}
$$
The ratio of the kinetic energies is now
$$
begin{align*}
frac{K_e}{K_p}&=frac{frac{h^2}{2m_e lambda^2}}{frac{h^2}{2m_p lambda^2}}\
&=frac{m_p}{m_e}
end{align*}
$$
The masses of the proton $m_p$ and the electron $m_p$ are
$$
begin{align*}
m_p=938frac{text{MeV}}{c^2}\
m_e=0.511frac{text{MeV}}{c^2}
end{align*}
$$
where $c$ is the speed of light in vacuum. Finally, the ratio of the kinetic energies of the proton and the electron (under the condition that they have the same de Broglie’s wavelengths) is
$$
begin{align*}
boxed{frac{K_e}{K_p}}&=frac{938frac{text{MeV}}{c^2}}{0.511frac{text{MeV}}{c^2}}\
&=boxed{1835.6}
end{align*}
$$
b) $frac{K_e}{K_p}=1835.6$
$m = 0.65$ kg
$lambda = 0.76$ m
a) Calculate the speed of the student to enter a window with $lambda = 0.76$ m:
$$
begin{align*}
lambda &= dfrac{h}{p} = dfrac{h}{mv} \
v &= dfrac{h}{m lambda} \
&= dfrac{6.626 times 10^{-34}}{(65) times (0.76)} \
&= boxed{1.34 times 10^{-35} text{ m/s}}
end{align*}
$$
$$
begin{align*}
text{time} &= dfrac{text{distance}}{text{speed}} \
&= dfrac{1.0 times 10^{-3} text{ m}}{1.34 times 10^{-35} text{ m/s}} \
&= boxed{7.5 times 10^{31} text{ s}}
end{align*}
$$
This time is longer than the age of the universe
b) $t = 7.5 times 10^{31}$ s
$$
begin{gather}
KE = dfrac{1}{2}mv^2 \
V = k Q/r \
PE = k q Q/r
end{gather}
$$
We can obtain the kinetic energy of the particle from the potential energy of $V$, such that:
$$
begin{align*}
KE &= PE \
&= k q Q/r
text{Using Eq(2) and Eq(3):} \
&= qV \
dfrac{1}{2}mv^2 &= qV \
text{We can obtain $v$ as follows} \
v &= sqrt{dfrac{2qV}{m}}
end{align*}
$$
Solving for the de Broglie wavelength:
$$
begin{align*}
lambda &= dfrac{h}{p} = dfrac{h}{mv} \
&= boxed{dfrac{h}{sqrt{2mqV}}}
end{align*}
$$
lambda = dfrac{h}{sqrt{2mqV}}
$$
b) Since the maximum kinetic energy will stay the same so will the maximum momentum and thus the minimum De Broglie Wavelength.
c) This will increase since there will be more photons and thus more electrons hit and ejected.
d) Since there will be more electrons hit the current that is comprised of them will increase.
b) Since the maximum kinetic energy increases so does the momentum and thus the minimum De Broglie wavelength will decrease (since it is inversely proportional to the momentum)y .
c) This would decrease since if we want to hold the intensity constant with more energetic photons we would have to decrease number of photons emitted per second and thus the number of electrons hit will decrease.
d) The current would remain the same since even though the ejected electrons would be faster there will be smaller number of them and the effects cancel out.
b) The best explanation literally from word to word is B.
a) Given what we know from part b) we see that since station A produces less energetic photons in order to radiate the same power it needs to compensate by producing more photons (since the emitted power is total energy emitted in unit of time).
$$
nu_{cutoff} = frac{W_0}{h}
$$
which yields for each material
$$
nu_{cuttoff_1} = frac{4.28times1.6times 10^{-19}text{ J}}{6.62times10^{-34}text{ J s}} =1.03times10^{15}text{ Hz},
$$
which is in ultraviolet part.
$$
nu_{cuttoff_1} = frac{4.25times1.6times 10^{-19}text{ J}}{6.62times10^{-34}text{ J s}} =1.03times10^{15}text{ Hz},
$$
which is also in ultraviolet part.
$$
nu_{cuttoff_1} = frac{2.14times1.6times 10^{-19}text{ J}}{6.62times10^{-34}text{ J s}} =5.17times10^{14}text{ Hz},
$$
which is the frequency of visible yellow light so you should use cesium.
$lambda = 545 times 10^{-9}$ m
$n = 100$ photons in a second
$textbf{Find}$ The power, $P$ delivered by the beam of light
$textbf{Formula:}$
$$
begin{gather}
P = E/t \
E = nhf = dfrac{nhc}{lambda}
end{gather}
$$
Using Eq (1) and Eq (2), we can solve for $P$ as follows:
$$
begin{align*}
text{We first obtain } E text{ as follows:} \
E &= dfrac{100 times 6.626 times 10^{-34} times 3.0 times 10^8}{545 times 10^{-9}} \
&= 3.65 times 10^{-17} text{ J} \
text{Calculating for the power we have:} \
P &= dfrac{E}{t} \
&= dfrac{3.65 times 10^{-17} text{ J}}{1 text{ s}} \
&= boxed{3.65 times 10^{-17} text{ W}}
end{align*}
$$
P = 3.65 times 10^{-17} text{ W}
$$
$$
P=frac{E}{1text{ s}} = frac{100hc/lambda}{1text{ s}} = frac{100times6.62times10^{-34}text{ J s}times 3times10^8text{ m/s}}{545times10^{-9}text{ m}times 1text{ s}} =3.64times10^{-17}text{ W}.
$$
$$
E=1.0times10^{-10}text{ J}
$$
in one second all comprised of photons. Thus we can write
$$
E=nhnu
$$
giving
$$
n=frac{E}{hnu} = frac{1.00times10^{-10}text{ J}}{6.62times10^{-34}text{ J s}times 96times 10^6text{ Hz}} = 1.57times10^{15}
$$
photons per second.
$$
nhnu=lambda_m m
$$
yielding
$$
n=frac{lambda_m m}{hnu} = frac{33.5times10^4text{ J/kg}times1.0text{ kg}}{6.62times10^{-34}text{ J s}times 6.0times10^{14}text{ Hz}} =8.43times10^{ 23}.
$$
The value of specific heat capacity of water is $4.184:J$.
Energy of a photon of wavelength 550 nm is given by
$E = dfrac{hc}{lambda} = dfrac{6.626 times 10^{-34} times 3 times 10^8}{550 times 10^{-9}} = dfrac{6.626 times 3}{550} times 10^{-34} times 10^8 times 10^9$
$E = 0.0361 times 10^{-17} = 3.61 times 10^{-19}:J$
So the number of photons required to increase the temperature of $1:g$ of water by $1^o:C$ is
$n = dfrac{4.184}{3.61} times 10^{19} = 1.159 times 10^{19} = 1159 times 10^{16}$.
So energy of the photon is given by
$E = hf = 6.626 times 10^{-34} times 2.11 times 10^{15} =13.98 times 10^{-19}:J$
The work function of the lead is given as
$W_0 = 4.25:eV = 4.25 :cancel{eV} times dfrac{1.6 times 10^{-19}:J}{1:cancel{eV}} =6.8 times 10^{-19}:J$
So the maximum kinetic energy of the electron is given by
$K_{max} = E – W_0 = (13.98 – 6.8) times 10^{-19} = 7.18 times 10^{-19}:J$
The maximum momentum of the electron is given by
$p_{max} = sqrt{2 mK_{max}} = sqrt{2 times 9.11 times 10^{-31} times 7.18 times 10^{-19}}$
$p_{max} = sqrt{2 times 9.11 times 7.18} times 10^{-25} = 11.44 times 10^{-25}:kg:m/s$
So the minimum de Broglie wavelength ($lambda_{min}$) of the ejected electrons is given by
$lambda_{min} = dfrac{h}{p_{max}} = dfrac{6.626 times 10^{-34}}{11.44 times 10^{-25}} = dfrac{6.626}{11.44} times 10^{-9} =0.579 times 10^{-9}:m = 0.579:nm$
lambda_{min} = 0.579:nm
$$
Mass of the electron is $m_e = 9.11 times 10^{-31}:kg$.
So momentum of the electron is given by
$p = m_ev_e = 9.11 times 10^{-31} times 2.7 times 10^6 = 9.11 times 2.7 times 10^{-25} = 24.597 times 10^{-25} : kg:m/s$
(a) The de Broglie wavelength of the electron is given by
$lambda = dfrac{h}{p} = dfrac{6.626 times 10^{-34}}{24.597 times 10^{-25}} = dfrac{6.626}{24.597} times 10^{-9} = 0.269 times 10^{-9}:m = 0.269:nm$
$m_p = 1.67 times 10^{-27}:kg$ is the mass of the proton.
(b) The momentum of the proton is given as same as that of electron and the de Broglie wavelength of the proton is same as that of electron.
(c) The speed of the proton is given by
$v_p = dfrac{p}{m_p} = dfrac{24.597 times 10^{-25}}{1.67 times 10^{-27}} = dfrac{24.597}{1.67} times 10^2 = 14.73 times 10^2 :m/s$
Since the momentum of proton and electron is same we can verify that
$m_e v_e = m_p v_p$ or
$dfrac{v_e}{v_p} = dfrac{m_p}{m_e}$
$dfrac{v_e}{v_p} = dfrac{2.7 times 10^6}{14.73 times 10^2} = dfrac{2.7}{14.73} times 10^4 = 1833$
$dfrac{m_p}{m_e} = dfrac{1.67 times 10^{-27}}{9.11 times 10^{-31}} = dfrac{1.67}{9.11} times 10^4 = 1833$
Calculation is verified.
$f_{peak} = 5.4 times 10^{14}$ Hz
$textbf{Find:}$ Temperature, $T$ of the blackbody
We calculate for the temperature of the blackbody as follows:
$$
begin{align*}
T &= dfrac{f_{peak}}{5.88 times 10^{10} text{ Hz/K}} \
&= dfrac{5.4 times 10^{14} text{ Hz}}{5.88 times 10^{10} text{ Hz/K}} \
&= boxed{9183 text{ K}}
end{align*}
$$
Based from the result, we cannot determine the firefly radiation by using blackbody radiation since the value obtained is very high, comparing this to the surface temperature of the Sun which is 6000 K this is not a good approximate value.
We know that the frequency of a simple pendulum $f$ of length $l$ is given by
$$
begin{align}
f=frac{1}{2pi}sqrt{frac{g}{l}}
end{align}
$$
where $g=9.81 frac{text{m}}{text{s}^2}$.
It is assumed that the energy of the system $E$ is given by
$$
begin{align}
E=nhf
end{align}
$$
where $n$ is the principal quantum number and $h=6.62times 10^{34}text{Js}$ is Planck’s constant.
The length of the string $l$ is:
$$
begin{align*}
l=0.78text{m}
end{align*}
$$
The mass of the object $m$ which is attached to the string is:
$$
begin{align*}
m=0.15text{kg}
end{align*}
$$
The principal quantum number $n$ is:
$$
begin{align*}
n=10^{33}
end{align*}
$$
a) We calculate the numerical value of the frequency of the pendulum $f$ by simply plugging in the numerical values for the length of the string $l$ and the mass of the object $m$:
$$
begin{align*}
boxed{f}&=frac{1}{2pi}sqrt{frac{g}{l}}\
&=frac{1}{2pi}sqrt{frac{9.81frac{text{m}}{text{s}^2}}{0.78text{m}}}\
&=boxed{0.56text{Hz}}
end{align*}
$$
b) In this part, it is required of us to calculate the maximum speed of the object $v_m$. The condition that the system satisfies in order for the object to have maximum speed is that the potential energy of the system is zero. Then, the total energy of the system $E$ is equal to the kinetic energy of the system $K$
$$
begin{align*}
E=K=frac{mv_m^2}{2}
end{align*}
$$
The maximum speed of the object $v_m$ is now
$$
begin{align*}
v_m=sqrt{frac{2E}{m}}
end{align*}
$$
If we use Eq. (2), we can calculate the maximum speed of the object $v_m$
$$
begin{align*}
boxed{v_m}&=sqrt{frac{2E}{m}}\
&=sqrt{frac{2nhf}{m}}\
&=sqrt{frac{2times 10^{33}times 0.56text{Hz}times 6.62 times 10^{34}text{Js}}{0.15text{kg}}}\
&=boxed{2.2frac{text{m}}{text{s}}}
end{align*}
$$
b) $v_m=2.2frac{text{m}}{text{s}}$
So energy of the photon is given by
$E = dfrac{hc}{lambda} = dfrac{6.626 times 10^{-34} times 3 times 10^8}{545 times 10^{-9}} = dfrac{6.626 times 3}{545} times 10^{-34} times 10^8 times 10^9$
$E =0.0365 times 10^{-17}:J = 3.65 times 10^{-19}:J$
The maximum speed of the emitted electrons is $v_{max} = 3.10 times 10^5 :m/s$
The maximum kinetic energy of the ejected electrons is given by
$K_{max} = dfrac{1}{2} mv_{max}^2 = dfrac{1}{2} times 9.11 times 10^{-31} times (3.10 times 10^5)^2$
where $m = 9.11 times 10^{-31}:Kg$ is the mass of the electron.
$= dfrac{9.11 times 3.10^2}{2} times 10^{-31} times 10^{10} = dfrac{9.11 times 9.61}{2} times 10^{-21}$
$K_{max}= 43.77 times 10^{-21} = 0.438 times 10^{-19}:J$
The work function of the metal is given by
$W_0 = E – K_{max} = 3.65 times 10^{-19} – 0.438 times 10^{-19} = (3.65 – 0.438) times 10^{-19}$
$W_0 = 3.212 times 10^{-19}:J$
At cutoff frequency the kinetic energy of the electrons is zero. So we have,
$h nu_0 = W_0$
So the cutoff frequency is given by
$nu_0 = dfrac{W_0}{h} = dfrac{3.212 times 10^{-19}}{6.626 times 10^{-34}} = dfrac{3.212}{6.626} times 10^{-19} times 10^{34} = 0.485 times 10^{15}:Hz$
$nu_0 = 0.485 times 10^{15}:Hz$
$E = dfrac{hc}{lambda}$
where $h= 6.626 times 10^{-34}:J.s$ is Planck’s constant and $c=3 times 10^8:m/s$ is the speed of light.
Hydrogen atom absorbs a photon of wavelength
$lambda = 486.2 :nm = 4862 times 10^{-10}:m$
So the energy of the absorbed photon is
$E_{abs} = dfrac{6.626 times 10^{-34} times 3 times 10^8}{4862 times 10^{-10}} = dfrac{6.626 times 3}{4862} times 10^{-34} times 10^8 times 10^{10}$
$E_{abs} =0.004088 times 10^{-16} = 0.4088 times 10^{-18}:J$
After short time it emits a photon of wavelength
$lambda = 97.23 :nm = 9723 times 10^{-11}:m$
So the energy of the emitted photon is
$E_{emi} = dfrac{6.626 times 3}{9723} times 10^{-34} times 10^8 times 10^{11} = 2.044 times 10^{-18}:J$
(b) The change in the energy of the hydrogen atom is the difference between energy of the absorbed and emitted photons. That is,
$Delta E = E_{abs} – E_{emi} = 0.4088 times 10^{-18} – 2.044 times 10^{-18} = (0.4088 – 2.044) times 10^{-18}$
$Delta E = -1.6352 times 10^{-18}:J$
$|Delta E| = 1.6352 times 10^{-18} :J$
$lambda = dfrac{h}{mv} = dfrac{h}{p}$ [$p=mv$ is the momentum of the particle.]
The kinetic of the particle is given by
$K = dfrac{p^2}{2m}$
This can be rewritten for momentum $p$, by multiplying both sides by $2m$ and taking square root on both sides, as
$p = sqrt{2mK}$
The de Broglie wavelength can be written in terms of kinetic energy as
$lambda = dfrac{h}{sqrt{2mK}}$
$lambda propto dfrac{1}{sqrt{K}}$
So the de Broglie wavelength of the particle decreases as the kinetic energy of the particle increases.
The value of Planck’s constant $h = 6.626 times 10^{-34}:J.s$
$1:eV = 1.6 times 10^{-19}:J$
So, $h = 6.626 times 10^{-34} :cancel{J}.s times dfrac{1:eV}{1.6 times 10^{-19} :cancel{J}} = dfrac{6.626}{1.6} times 10^{-34} times 10^{19}$
$h = 4.141 times 10^{-15}:eV$
So the de Broglie wavelength of the electron can be written as
$lambda = dfrac{4.141 times 10^{-15}}{sqrt{2 times 9.11 times 10^{-31} times K}} = dfrac{4.141 times cancel{10^{-15}}}{sqrt{2 times 91.1} times cancel{10^{-15}} times sqrt{K}} = dfrac{4.141}{13.5 sqrt{K}}$
$lambda = dfrac{0.307}{sqrt{K}}$
where the kinetic energy is in the unit of joule or J. Converting this into eV.
$lambda = dfrac{0.307}{sqrt{dfrac{K}{1.6 times 10^{-19}}}}= dfrac{0.307 times sqrt{1.6 times 10^{-19}}}{sqrt{K}}= dfrac{0.307 times 4 times 10^{-9}}{sqrt{K}} =dfrac{1.23:nm}{sqrt{K}}$
$hf = KE_{{max}_A} + W_A$ (for surface $A$)
$hf = KE_{{max}_B} + W_B$ (for surface $B$)
where $KE_{{max}_A}$ and $KE_{{max}_B}$ are the maximum kinetic energy of the electrons emitted from the surface $A$ and $B$ respectively.
Subtracting these equations we have,
$0 = (KE_{{max}_A} – KE_{{max}_B}) + (W_A – W_B)$
or
$KE_{{max}_A} – KE_{{max}_B} = W_B – W_A$
(a) The difference in the maximum kinetic energies of the electrons emitted from two surfaces is independent of the frequency of the incident light. It depends only on the difference in their work function.
So when frequency of the incident light is increased, the difference in the maximum kinetic energies remains same.
(b) Electrons emitted from the surface $A$ has maximum kinetic energy that is $2.00 times 10^{-19}:J$ more than the maximum kinetic energy of the electrons emitted from surface $B$. That is,
$KE_{{max}_A} = 2.00 times 10^{-19}:J + KE_{{max}_B}$
So, $W_B – W_A = KE_{{max}_A} – KE_{{max}_B} = 2.00 times 10^{-19}:J + KE_{{max}_B} – KE_{{max}_B}$
$W_B – W_A = 2.00 times 10^{-19}:J$
$$
begin{align*}
380text{nm} text{to} 700text{nm}
end{align*}
$$
On the other hand, electrons in the electron microscopes can have wavelengths as low as
$$
begin{align*}
0.01text{nm}
end{align*}
$$
We see that the wavelengths associated with visible light are $1000$ times greater than the wavelengths associated with electrons. We conclude that electron microscopes have higher resolving power than light microscopes and can reveal the structure of smaller objects.
If the concept of wave-particle was not involved, we couldn’t even talk about the wavelength of an electron.
$lambda_1 = 433.9 times 10^{-9}$ m
$KE_{max}^{1} = 0.550$ eV
$lambda_2 = 253.5 times 10^{-9}$ m
$KE_{max}^{2} = 2.57$ eV
$textbf{Find:}$ $W_o$
To obtain $W_o$, we first set-up the equations for the two experiments as follows:
$$
begin{align*}
KE_{max} &= dfrac{hc}{lambda} – W_o \
W_o &= dfrac{hc}{lambda} – KE_{max} \
text{for the first experiment we have:} \
W_o &= dfrac{h times (3.0 times 10^8 text{ m/s})}{433.9 times 10^{-9} text{ m}} – 0.550 text{ eV} \
text{for the secondexperiment we have:} \
W_o &= dfrac{h times (3.0 times 10^8 text{ m/s})}{253.5 times 10^{-9} text{ m}} – 2.57 text{ eV} \
end{align*}
$$
$$
begin{align*}
text{using the equation for the first element we have:} \
h &= dfrac{(W_o + 0.550) times (433.9 times 10^{-9})}{3.0 times 10^8} \
text{substituting this to h in the equation for the second experiment we have:} \
W_o &= dfrac{(dfrac{(W_o + 0.550) times (433.9 times 10^{-9})}{3.0 times 10^8} times (3.0 times 10^8 text{ m/s})}{253.5 times 10^{-9} text{ m}} – 2.57 text{ eV} \
end{align*}
$$
begin{align*}
text{converting eV to J we have:} \
W_o &= dfrac{(dfrac{(W_o + 8.8 times 10^{-20}) times (433.9 times 10^{-9})}{3.0 times 10^8} times (3.0 times 10^8 text{ m/s})}{253.5 times 10^{-9} text{ m}} – 4.1 times 10^{-19} text{ J} \
text{solving for } W_o\
&= 3.64 times 10^{-19} text{ J}\
text{converting it to eV we have:} \
&= 3.64 times 10^{-19} times dfrac{1 text{ eV}}{1.6 times 10^{-19}} \
&= boxed{2.29 text{ eV}}
end{align*}
$$
$lambda_1 = 433.9 times 10^{-9}$ m
$KE_{max}^{1} = 0.550$ eV
$lambda_2 = 253.5 times 10^{-9}$ m
$KE_{max}^{2} = 2.57$ eV
$W_o = 2.29$ eV (obtained from Problem 100)
$textbf{Find:}$ $h$
To obtain $h$, we first set-up the equations for the two experiments as follows:
$$
begin{align*}
KE_{max} &= dfrac{hc}{lambda} – W_o \
h &= dfrac{(KE_{max} + W_o) times lambda}{c} \
text{for the first experiment we have:} \
h &= dfrac{(8.8 times 10^{-20} + 3.64 times 10^{-19}) times 433.9 times 10^{-9}}{3.0 times 10^8} \
&= boxed{6.57 times 10^{-34} text{ Js}}
end{align*}
$$
Einstein’s formula for the photoelectric effect is given by
$$
begin{align}
hf=K+W_0
end{align}
$$
where $K$ is the kinetic energy of the ejected electron, $W_0$ is the work function for the metal, $f$ is the frequency of the incident photon and $h=6.62times 10^{-34}text{Js}$ is Planck’s constant.
The wavelength of the light $lambda$ that Millikan used is:
$$
begin{align*}
lambda=365text{nm}
end{align*}
$$
The work function of lithium $W_0$ is
$$
begin{align*}
W_0&=2.29text{eV}\
&=2.29times 1.6 times 10^{-19}text{J}=3.7times 10^{-19}text{J}
end{align*}
$$
The relationship between the frequency $f$ and the wavelength $lambda$ is given by
$$
begin{align*}
f=frac{c}{lambda}
end{align*}
$$
where $c=3times 10^8 frac{text{m}}{text{s}}$ is the speed of light in vacuum. From Eq. (1) we can express the kinetic energy $K$ and plug in the given numerical values
$$
begin{align*}
boxed{K}&=frac{hc}{lambda}-W_0\
&=frac{6.62times 10^{-34}text{Js}times 3 times 10^8frac{text{m}}{{s}}}{365times 10^{-9}text{m}}-3.7times 10^{-19}text{J}\
&=1.3times 10^{-19}text{J}\
&=1.3times 10^{-19} times 6.24 times 10^{18}text{eV}\
&=boxed{0.805text{eV}}
end{align*}
$$
