All Solutions
Page 906: Assessment
In Rutherford’s experiment, on the other hand, it was shown that only some (a little fraction) alpha particles were ‘deflected’ by the gold foil. Thus, it was concluded that the positively charged particles are concentrated in a small region of the atom, and were not scattered around the atom, like in the plum-pudding model.
The emission spectra of a chemical element or chemical compound is the spectrum of frequencies of electromagnetic radiation emitted due to an atom or molecule making a transition from a high energy state to a lower energy state. Each element’s emission spectrum is unique.
$$
begin{gather*}
d_{nucleus} = 1.0 times 10^{-15} text{ m} \
r_{atom} = 5.3 times 10^{-11} text{ m}
end{gather*}
$$
The volume of a sphere can be calculated using the formula:
$$
begin{gather}
V = dfrac{4}{3} pi r^3
end{gather}
$$
We use Eq (1) to be able to obtain the $v_{nucleus}$ and $v_{atom}$, such that:
$$
begin{align*}
v_{nucleus} &= dfrac{4}{3} left(dfrac{d_{nucleus}}{2}right)^3 pi \
&= dfrac{4}{3} left(dfrac{1.0 times 10^{-15} text{ m}}{2}right)^3 pi \
v_{atom} &= dfrac{4}{3} (5.3 times 10^{-11} text{ m})^3 pi \
end{align*}
$$
Using the values for $v_{nuclues}$ and $v_{atom}$, we can get the fraction occupied by the nucleus as follows:
$$
begin{align*}
text{fraction occupied by nucleus} &= dfrac{v_{nucleus}}{v_{atom}} \
&= dfrac{dfrac{4}{3} left(dfrac{1.0 times 10^{-15} text{ m}}{2}right)^3 pi}{dfrac{4}{3} (5.3 times 10^{-11} text{ m})^3 pi } \
&= 8.4 times 10^{-16}
end{align*}
$$
Hence, the fraction of the atom occupied by the nucleus is $boxed{8.4 times 10^{-16}}$
diameter$_{baseball} = 7.3$ cm or $0.073$ m
diameter$_{nucleus} = 1.0 times 10^{-15}$ m
distance$_{atom} = 5.3 times 10^{-11}$ m
To be able to visualize the size of the atom as the size of a baseball, we get the proportions of the atom and baseball such that:
$$
begin{gather}
dfrac{text{distance}_{baseball}}{text{diameter}_{baseball}} = dfrac{text{distance}_{atom}}{text{diameter}_{nucleus}}
end{gather}
$$
Using Eq (1), we can get the distance from the center of the baseball as follows:
$$
begin{align*}
text{distance}_{baseball} &= dfrac{text{distance}_{atom}}{text{diameter}_{nucleus}} times text{diameter}_{baseball} \
&= dfrac{5.3 times 10^{-11} text{ m}}{1.0 times 10^{-15} text{ m}} times 0.073 text{ m} \
&= 3869 text{ m}
end{align*}
$$
Hence, if the nucleus is of the size of a baseball, the distance from the center will be 3869 m
$$
frac{r_{atom}}{d_{nucleus}} = frac{text{Radius}}{text{Diameter}}.
$$
This will give
$$
text{Radius} = frac{r_{atom}}{d_{nucleus}}text{Diameter} = frac{5.3times10^{-11}text{ m}}{1.00times10^{-15}text{ m}}0.073text{ m} approx 3900text{ m}
$$
so the typical distance would be of the order of a few kilometers.
$$
a = frac{v^2}{r}
$$
. The electrostatic attractive force would be
$$
F=kfrac{text{electron charge}timestext{proton charge}}{r^2} = kfrac{e^2}{r^2}.
$$
Now equating as required in the problem we get
$$
kfrac{e^2}{r^2} = mfrac{v^2}{r}Rightarrow v^2 = frac{ke^2}{rm}Rightarrow v=sqrt{frac{ke^2}{rm}} =sqrt{frac{9times 10^9text{Nm}^2text{C}^2e^2}{rm}} =2.18times10^6text{ m/s}.
$$
$$
PE = kfrac{text{ charge of the alpha particle }times text{ charge of gold atom nucleus }}{r}=kfrac{158e^2}{r}
$$
where $r$ is the required radius. Equating this to particle’s kinetic energy as instructed we get
$$
KE=kfrac{158e^2}{r}
$$
giving
$$
r=frac{k 158e^2}{KE} = frac{9times10^{9}text{ Nm}^2text{/C}^2times 158times(1.6times 10^{-19}text{ C})^2}{3times10^61.6times10^{-19}text{ J}} =7.6times 10^{-14}text{ m}.
$$
b) There is no upper limit. The difference in energies between neighboring orbits with arbitrarily high $n$ is arbitrarily small (since the energy is inversely proportional to $n^2$ ) and thus $lambda$ can become arbitrarily large.
$$
r_1=lambda_1=(5.29times10^{-11}text{ m})1^2quad r_3 = 3lambda_3 = (5.29times10^{-11}text{ m})3^2
$$
Dividing these equations we get
$$
frac{3lambda_3}{lambda_1} = 9
$$
since everything cancels out and finally we have
$$
lambda_3 = 3lambda_1.
$$
$n_{1}=4$
$$
n_{2}=infty
$$
Unknown:
$$
Delta E=?
$$
Solution:
$$
Delta E=left(-13.6 {rm eV}right)left(frac{1}{n_{2}^{2}}-frac{1}{n_{1}^{2}}right)=left(-13.6 {rm ev}right)left(frac{1}{infty}-frac{1}{4^{2}}right)=0.85 {rm eV}
$$
textbf{Known:}
$$
$n_{1}=2$
$$
n_{2}=5
$$
$$
textbf{Unknown:}
$$
$$
Delta E=?
$$
$$
textbf{Solution:}
$$
$$
Delta E=left(-13.6 {rm eV}right)left(frac{1}{n_{2}^{2}}-frac{1}{n_{1}^{2}}right)=left(-13.6 {rm ev}right)left(frac{1}{5^{2}}-frac{1}{2^{2}}right)=2.86 {rm eV}
$$
$$
begin{align*}
& nlambda=2pi r_{n}\
Rightarrow & lambda=frac{2pi r_{n}}{n}
end{align*}
$$
Now putting $n=1$, we have
$$
lambda=frac{2pileft(5.22times10^{-11} {rm m}right)}{1}=3.28times10^{-10} {rm m}
$$
3.28times10^{-10} {rm m}
$$
the length of orbit. Hence the quantum number is 5. Now for hydrogen
atom the radius of the Bohr orbit is given by
$$
begin{align*}
r_{n} & =left(5.22times10^{-11} {rm m}right)n^{2}\
& =left(5.22times10^{-11} {rm m}right)5^{2}\
& =1.30times10^{-9} {rm m}
end{align*}
$$
1.30times10^{-9} {rm m}
$$
[
Delta E=left(-13.6 {rm eV}right)left(frac{1}{n_{1}^{2}}-frac{1}{n_{2}^{2}}right)
]
We have $n_{1}=6$ and $n_{2}=5$, hence
[
Delta E=left(-13.6 {rm eV}right)left(frac{1}{6^{2}}-frac{1}{5^{2}}right)=0.166 {rm eV}
]
So the wavelength of the light is
[
lambda=frac{hc}{Delta E}=frac{left(4.13times10^{-15} {rm eVcdot s}right)left(3times10^{8} {rm m/s}right)}{left(0.166 {rm eV}right)}=7.46times10^{-6} {rm m}=7.46 mbox{textmu m}
]
So the orbital radius is
$$
r_{n}=left(5.22times10^{-11} {rm m}right)n^{2}=left(5.22times10^{-11} {rm m}right)8^{2}=3.34times10^{-9} {rm m}
$$
$$
E_{n}=left(-13.6 {rm eV}right)frac{1}{n^{2}}=left(-13.6 {rm eV}right)frac{1}{8^{2}}=0.213 {rm eV}
$$
Now we have
$$
begin{align*}
& E_{n}=left(-13.6 {rm eV}right)frac{1}{n^{2}}\
Rightarrow & n=sqrt{frac{left(-13.6 {rm eV}right)}{E_{n}}}=sqrt{frac{-13.6 {rm e}V}{-1.51 {rm eV}}}=3
end{align*}
$$
Hence the orbit of the Bohr state is
$$
r_{n}=left(5.22times10^{-11} {rm m}right)n^{2}=left(5.22times10^{-11} {rm m}right)3^{2}=4.70times10^{-10} {rm m}
$$
4.70times10^{-10} {rm m}
$$
$$
r_{n}=left(5.22times10^{-11} {rm m}right)n^{2}
$$
Therefore we have
$$
n=sqrt{frac{r_{n}}{5.22times10^{-11} {rm m}}}=sqrt{frac{1.32times10^{-9} {rm m}}{5.22times10^{-11} {rm m}}}=5.02approx5
$$
Hence the energy is given by
$$
E_{n}=left(-13.6 {rm eV}right)frac{1}{n^{2}}=left(-13.6 {rm eV}right)frac{1}{5^{2}}=0.544 {rm eV}
$$
$$
n_{1}=3
$$
$Delta E=1.23$ eV
Unknown:
$n_{2}=?$
Solution:
We know that
$$
begin{align*}
& Delta E=left(-13.6 {rm eV}right)left(frac{1}{n_{2}^{2}}-frac{1}{n_{1}^{2}}right)\
Rightarrow & frac{1}{n_{2}^{2}}=frac{Delta E}{left(-13.6 {rm eV}right)}+frac{1}{n_{1}^{2}}=frac{1.23 {rm eV}}{left(-13.6 {rm eV}right)}+frac{1}{3^{2}}=0.0207\
Rightarrow & n_{2}=sqrt{frac{1}{0.0207}}=6.95approx7
end{align*}
$$
So the $n$ value for the final state is 7.
n=7
$$
$lambda = 656$ nm
We are to find the the initial and final states of the electron in the hydrogen that emits this wavelength.
Formula:
$$
begin{gather*}
lambda = dfrac{hc}{|Delta E|} \
end{gather*}
$$
To find the initial and final states, we first have to find the change in energy ($Delta E$) that was produced in this transition, such that:
$$
begin{align*}
|Delta E | &= dfrac{hc}{lambda} \
&= dfrac{(6.626 times 10^-{34} text{J s} ) times (3.0 times 10^8 text{m/s)}}{656 times 10^{-9} text{m}} \
&= dfrac{(6.626 times 10^{-34} text{J s} ) times left (dfrac{1 text{eV}}{1.60 times 10^{-19} text{J}} right ) times (3.0 times 10^8 text{m/s)}}{656 times 10^{-9} text{m}} \
&= 1.89 text{eV}
end{align*}
$$
We know that the wavelength $656$ nm is between $600-700$ nm, which is where the red spectral line of hydrogen forms. The red spectral line of hydrogen corresponds to a quantum jump from n = $3$ to n = $2$. Let’s check if the initial and final states also correspond to these values:
$$
begin{gather*}
E_2 = (-13.6 text{eV}) times dfrac{1}{2^2} = -3.4 text{eV} \
E_3 = (-13.6 text{eV}) times dfrac{1}{3^3} = -1.51 text{eV}
end{gather*}
$$
$$
begin{align*}
|Delta E | &= |(E_2 – E_3)| \
&= |-3.4 text{ eV} – (- 1.51 text{ eV})| \
&= 1.89 text{ eV}
end{align*}
$$
Hence, our initial and final states are $boxed{ n = 3 text{ and } n = 2}$, respectively
$$
Delta E=left(-13.6 {rm eV}right)left(frac{1}{n_{2}^{2}}-frac{1}{n_{1}^{2}}right)=left(-13.6 {rm eV}right)left(frac{1}{5^{2}}-frac{1}{3^{2}}right)=0.967 {rm eV}
$$
$$
Delta E=left(-13.6 {rm eV}right)left(frac{1}{7^{2}}-frac{1}{5^{2}}right)=0.266 {rm eV}
$$
$n_{1}=7,$ $n_{2}=8$
The shortest wavelength will be when the energy is maximum, that is
the case B. That is when $n_{1}=2, n_{2}=8$
For case C, wavelength is
$$
lambda=frac{hc}{Delta E}=frac{hc}{left(-13.6 {rm eV}right)left(frac{1}{n_{2}^{2}}-frac{1}{n_{1}^{2}}right)}=frac{left(4.13 times10^{-15} {rm eV}cdot sright)left(3times10^{8} {rm m/s}right)}{left(-13.6 {rm eV}right)left(frac{1}{8^{2}}-frac{1}{7^{2}}right)}=1.90times10^{-5} {rm m}
$$
For case B, the wavelength is
$$
lambda=frac{hc}{Delta E}=frac{hc}{left(-13.6 {rm eV}right)left(frac{1}{n_{2}^{2}}-frac{1}{n_{1}^{2}}right)}=frac{left(4.13 times10^{-15} {rm eV}cdot sright)left(3times10^{8} {rm m/s}right)}{left(-13.6 {rm eV}right)left(frac{1}{8^{2}}-frac{1}{2^{2}}right)}=3.89times10^{-7} {rm m}
$$
The potential energy $U$ of a hydrogen atom is given by
$$
begin{align}
U=-frac{1}{4pi epsilon_0}frac{e^2}{r}
end{align}
$$
where $e=-1.6times 10^{-19}text{C}$ is the charge of the electron and $r$ is the distance of the electron from the nucleus, or, in terms of Bohr’s theory, it is the radius of a particular orbit.
The radius of a particular orbit $r_n$ is given by
$$
begin{align}
r_n=(5.2times 10^{-10}text{m})n^2
end{align}
$$
where $n$ is the principal quantum number.
The numerical value of the potential energy $U$ is:
$$
begin{align*}
U=-1.2times 10^{-19}text{J}
end{align*}
$$
a) We can calculate the radius of the desired Bohr orbit $r$ by expressing it in terms of the potential energy $U$. From Eq. (1) we get
$$
begin{align*}
r_n=-frac{1}{4pi epsilon_0}frac{e^2}{U}\
&=-frac{1}{9times 10^9frac{text{Jm}}{text{C}^2}}frac{(1.6times 10^{-19}text{C})^2}{-1.2times 10^{-19}text{C}}\
&=1.9times 10^{-9}text{m}
end{align*}
$$
Now, we can calculate the Bohr’s orbit, ie. the principal quantum number $n$ that corresponds to the calculated radius of the orbit. We do this by expressing $n$ from Eq. (2). We get
$$
begin{align*}
boxed{n}&=sqrt{frac{r_n}{5.2times 10^{-10}text{m}}}\
&=sqrt{frac{1.9times 10^{-9}text{m}}{5.2times 10^{-10}text{m}}}\
&=boxed{6}
end{align*}
$$
b) If the electron jumps to a higher Bohr orbit, the value of the principal quantum number $n$ increases. This means that the radius of the orbit also increases (it is easily seen from Eq. (2)). Now, if the radius $r$ increases, we conclude that the absolute value of the potential energy $|U|$ decreases (see the dependance of the potential energy $U$ on the distance $r$ in Eq. (1)).
b) The absolute value of the potential energy $|U|$ will decrease when the electron jumps to a higher Bohr orbit.
2) Intensity of laser light is higher than normal light.
3) Highly directional, that is like other light, laser light beam diameter does not increase so fast.
4) Laser light are highly monochromatic.
5) In general laser light are polarized.
$lambda = 193$ nm
Find: $Delta$ E
Formula:
$|Delta E|$= $dfrac{hc}{lambda}$
$$
begin{align*}
|Delta E| &= dfrac{(6.626 times 10^{-34} text{ J s} ) times (3.0 times 10^8 text{ m/s)}}{193 times 10^{-9} text{ m}} \
&= 1.03 times 10^{-18} text{ J}
end{align*}
$$
Energy difference between the two levels is $boxed{1.03 times 10^{-18} text{ J}}$
Energy delivered by the laser = $1.58 times 10^{-15}$ J
Find: Number of photons delivered
Solution:
$$
begin{align*}
text{number of photons} &= dfrac{E_{laser}}{Delta E} \
&= dfrac{1.58 times 10^{-13}}{1.03 times 10^{-18}} \
&= 1534
end{align*}
$$
Number of photons required to deliver the energy to the cornea is $boxed{1.53 times 10^5 text{photons}}$
b) $1.53 times 10^5$ photons
$$
begin{align}
triangle E=hnu=(-13.6text{eV}) left( frac{1}{n_f^2}-frac{1}{n_i^2} right)
end{align}
$$
where we supposed that the electron jumps from level $n_i$ to level $n_f$; here, $h$ is Planck’s constant, $nu$ is the frequency of the photon. In terms of the wavelength $lambda$, the equation above is written as
$$
begin{align}
frac{hc}{lambda}=(-13.6text{eV}) left( frac{1}{n_f^2}-frac{1}{n_i^2} right)
end{align}
$$
where $c$ is the speed of light in vacuum. We see that, when the difference between the two levels increases, the frequency of the photon $nu$ will also increase, and vice versa, the wavelength of the photon $lambda$ will decrease. Thus, in our case, we rank the jumps as:
a) B$<$A$<C$
b) C$<$A$<$B
b) C$<$A$<$B
$$
Delta E=(-13.6 eV)left(frac{1}{2^2}-frac{1}{1^2}right)=10.2 eV
$$
Hence the frequency of the photon is
$$
nu=frac{E}{h}=frac{13.6 eV}{4.13times 10^{-15} eVs}=3.29times 10^{-15} mathrm{Hz}
$$
3.29times 10^{-15} mathrm{Hz}
$$
Bohr radius ($r_n$) = $8.0$ microns or $8.0 times 10^{-6}$ m
Find: n
Formula:
$$
begin{align}
r_n = (5.22 times 10^{-11} m) times n^2
end{align}
$$
We need to get an equation that will help us solve for $n$, we can use Eq (1) such that:
$$
begin{align*}
n^2 &= dfrac{r_n}{5.22 times 10^{-11} text{m}} \
n &= sqrt{dfrac{8.0 times 10^{-6}}{5.22 times 10^{-11}}} \
&= 391.48
end{align*}
$$
Hence, the smallest value of $n$ such that the Bohr radius of a single hydrogen atom is greater than $8.0$ microns is $boxed{392}$
$n_{initial} = 4$
$n_{final} = 2$
Formula:
$$
begin{align}
p = dfrac{h}{lambda} \
|Delta E| = dfrac{h c}{lambda}
end{align}
$$
a) Find the momentum of the photon emitted
We cannot solve $p$ using only Eq (1) since we do not have a known value for $lambda$, using Eq (1) and Eq (2), we can find a relationship that will help us solve for the momentum, such that:
$$
begin{align*}
p &= dfrac{h}{lambda} \
&= dfrac{h c}{lambda c} \
&= dfrac{|Delta E|}{c}
end{align*}
$$
Solving for $|Delta E|$:
$$
begin{align*}
|Delta E| &= (-13.6 text{ eV}) times (dfrac{1}{4^2} – dfrac{1}{2^2}) \
&= 2.55 text{ eV}
end{align*}
$$
We can now solve for $p$, using our derived equation:
$$
begin{align*}
p &= dfrac{(2.55 text{ eV})left (dfrac{1.60 times 10^{-19} text{J}}{1 text{ eV}} right )} {3.0 times 10^8 text{ m/s}} \
&= 1.36 times 10^{-27} text{kg m/s}
end{align*}
$$
Momentum is $boxed{1.36 times 10^{-27} kg m/s}$
$$
begin{align*}
p &= text{mass of hydrogen atom} times text{velocity} \
text{velocity} &= dfrac{1.36 times 10^{-27} text{kg m/s}}{1.68 times 10^{-27} text{kg}} \
&= 0.81 text{ m/s}
end{align*}
$$
Hence, the recoil speed of the hydrogen atom is $boxed{0.81m/s}$
b) $v = 0.81 m/s$
$$
begin{gather}
F = qvB \
F_c = dfrac{mv^2}{r_n}
end{gather}
$$
Since the magnetic force makes circular motion happen, we have a relationship between magnetic force and centripetal force by combining Eq (1) and Eq (2):
$$
begin{align}
dfrac{mv^2}{r_n} &= qvB \
mv &= qBr_n
end{align}
$$
Suppose that the angular momentum is quantized, we also have the following equations:
$$
begin{gather*}
mvr_n = n times hslash \
n = 1, 2, 3, … \
hslash = dfrac{h}{2pi}
end{gather*}
$$
From these equations, we can find the radii of its allowed orbits, such that:
$$
begin{align*}
mvr_n &= n times hslash \
(qBr_n) r_n &= n times hslash \
r_{n}^{2} &= dfrac{n times hslash}{qB} \
r_n &= boxed{sqrt{dfrac{n times hslash}{qB}}}
end{align*}
$$
$$
begin{align*}
v &= dfrac{qBr_n}{m} \
&= dfrac{qB}{m} sqrt{dfrac{n times hslash}{qB}} \
&= boxed{dfrac{1}{m} sqrt{qBn times hslash}}
end{align*}
$$
b) $v = dfrac{1}{m} sqrt{qBn times hslash}$
Laser was invented by Charles Hard Townes and Arthur Leonard Schawlow.
There are different type of laser like:
* Gas laser,
* chemical laser,
* die laser,
* solid state laser,
* fiber laser,
* semiconductor laser, etc.
Usage:
* Laser is used for different kind of surgery.
* It is used in laser printer, cd drive etc.
* In research like spectroscopy, laser cooling, etc.
* In industry, like cutting metal, welding, etc.
1) The quantization: That is the electron can’t have continuous energy. The energy of a electron in atom is quantized.
2) The wave particle duality: That is the electron can be considered as particle and wave together.
3) Probability: It is not possible the find out exact location of electron, we can only calculate the probability of finding the electron at a particular position.
Because of the quantization the atomic transitions are discrete, and hence we get a lines of color instead of continuous spectrum.
The quantization of the energy states can be explained using wave particle duality. That is since electrons can also behave like wave, only in some orbit the length of the orbit is integer multiple of the wavelength. Hence on this particular orbit the electron makes standing wave and hence the corresponding state is an allowed state. Where as in other states the interference become destructive and state is not allowed.
The probability cloud in atoms give us the probability of finding the electron in the atom. This is some what equivalent to the classical orbit of the electron on the particular state.
number of photons emitted by argon laser = $1.49 times 10^{19}$ photons
$lambda_{1} = 488$ nm
$lambda_{2} = 514.5$ nm
Find:
Power output of laser (in watts)
We first solve for the number of photons with $lambda_1$ and $lambda_2$:
$$
begin{align*}
n &= dfrac{1.49 times 10^{19} text{photons}}{2} \
&= 7.45 times 10^{18} text{ photons per second for each } lambda
end{align*}
$$
Let’s solve for the corresponding power output for photons that have $lambda_1$ and $lambda_2$:
$$
begin{align*}
text{Power} &= dfrac{nhc}{lambda} \
P_1 &= dfrac{(7.45 times 10^{18} text{photons/s})(6.626 times 10^{-34} text{J s})(3 times 10^8 text{ m/s})}{488 times 10^{-9} text{ m}} = 3.04 text{W} \
P_2 &= dfrac{(7.45 times 10^{18})(6.626 times 10^{-34})(3 times 10^8)}{514.5 times 10^{-9}} = 2.88 text{W} \
text{Total Power} &= P_1 + P_2 \
&= 3.04 + 2.88 \
&= boxed{5.92 text{ W}}
end{align*}
$$
Wavelength $lambda=414.0 {rm nm}=414.0times10^{-9} {rm m}$
There fore the power is given by
$$
P=nfrac{hc}{lambda}=left(1.49times10^{19} {rm /s}right)frac{left(6.63times10^{-34} {rm Js}right)left(3times10^{8} {rm m/s}right)}{left(414.0times10^{-9} {rm m}right)}=7.16 {rm W}
$$
So the power is 7.16 W, the correct answer is D.
$$
Delta E=frac{hc}{lambda}=frac{left(4.13times10^{-15} {rm eVs}right)left(3times10^{8} {rm m/s}right)}{left(514.5times10^{-9} {rm m}right)}=2.42 {rm eV}
$$
So the correct answer is B.
