a) We can read the figure 27-1 where we see that the peak frequencies are

$$
f(4000textrm{ K})=2.5times 10^{14}textrm{ Hz}
$$

$$
f(5800textrm{ K})=3.5times 10^{14}textrm{ Hz}
$$

$$
f(8000textrm{ K})=4.5times 10^{14}textrm{ Hz}
$$

b) We see that the frequency of the peak intensity increases linearly with the temperature.

c) The intensity in the red part of the spectrum rises from 0.5 to 9 which is a 18-fold increase.

$$
textrm{a) }f(4000textrm{ K})=2.5times 10^{14}textrm{ Hz}
$$

$$
textrm{a) }f(5800textrm{ K})=3.5times 10^{14}textrm{ Hz}
$$

$$
textrm{a) }f(8000textrm{ K})=4.5times 10^{14}textrm{ Hz}
$$

$$
textrm{ b) It increases with the temperature.}
$$

$$
textrm{c) It increases 18 times.}
$$

If we take the given speed the and assume that a baseball weighs $0.2$kg we get that the de Broglie wavelength is given as

$$
lambda=frac{h}{mv}=frac{6.63times 10^{34}}{0.2times 21}
$$

which gives that

$$
lambdaapprox10^{-34}textrm{ m}
$$

Which is $10^{33}$ times less than the diameter of the ball.

The de Broglie wavelength of the ball is $10^{33}$ times less than its diameter.

According to Planck’s theory the energy is quantized as

$$
E=nhf
$$

from where we can express the frequency as

$$
f=frac{E}{nh}=frac{5.44times 10^{-19}}{1times 6.63times 10^{-34}}
$$

Finally, we have that

$$
boxed{f=8.21times 10^{14}textrm{ Hz}}
$$

$$
f=8.21times 10^{14}textrm{ Hz}
$$

When the electrons are stopped the kinetic energy is balanced out by the electrostatic potential completely therefore it has to hold

$$
K=-qV_0
$$

Therefore the stopping potential is given as

$$
V_0=-frac{K}{q}=frac{4.8times 10^{-19}}{1.6times 10^{-19}}
$$

Finally,

$$
boxed{V_0=3textrm{ V}}
$$

$$
V_0=3textrm{ V}
$$

According to quantum theory the momentum of photon is given as

$$
p=frac{h}{lambda}=frac{6.63times 10^{-34}}{4times 10^{-7}}
$$

Finally, one gets that

$$
boxed{p=1.7times 10^{-27}, frac{textrm{kg}cdottextrm{m}}{textrm{s}}}
$$

$$
p=1.7times 10^{-27}, frac{textrm{kg}cdottextrm{m}}{textrm{s}}
$$

In order to solve this problem we have to understand that the maximum kinetic energy of an ejected electron is equal to the stopping potential i.e.

$$
K_{max}=eV_0
$$

a) So in electron-volts we simply have that

$$
K_{max}=etimes5textrm{ V}
$$

$$
boxed{K_{max}=5textrm{ eV}}
$$

b) In joules, the same energy has the following value

$$
K_{max}=5times 1.6times 10^{-19}textrm{ J}
$$

$$
boxed{K_{max}=8times 10^{-19}textrm{ J}}
$$

$$
textrm{a) }K_{max}=5textrm{ eV}
$$

$$
textrm{b) }K_{max}=8times 10^{-19}textrm{ J}
$$

In order to solve this problem, we are going to use the photoelectric effect equation which tells us that the maximum kinetic energy is given as

$$
K_{max}=E_{ph}-W=hf-hf_0
$$

On the other hand the frequency of the light is given as

$$
f=frac{c}{lambda}=frac{3times 10^8}{650times 10^{-9}}
$$

$f=4.6times 10^{14}textrm{ Hz}$
so we have that

$$
K_{max}=h(f-f_0)=6.63times 10^{-34}times(4.6times 10^{14}-3times 10^{14})
$$

Finally, we have that

$$
boxed{K_{max}=1.06times 10^{-19}textrm{ J}}
$$

$$
K_{max}=1.06times 10^{-19}textrm{ J}
$$

By definition, the work needed to be done in order to free an electron from the surface is related to the threshold frequency as follows

$$
W=hf_0=6.63times 10^{-34}times 4.4times 10^{14}
$$

which gives that

$$
boxed{W=2.9times 10^{-19}textrm{ J}}
$$

$$
W=2.9times 10^{-19}textrm{ J}
$$

In order to solve this problem, we are going to use the photoelectric effect equation which tells us that the maximum kinetic energy is given as

$$
K_{max}=E_{ph}-W=hf-hf_0
$$

$$
K_{max}=h(f-f_0)=6.63times 10^{-34}times(1times 10^{15}-4.4times 10^{14})
$$

Finally, we have that

$$
boxed{K_{max}=3.7times 10^{-19}textrm{ J}}
$$

$$
K_{max}=3.7times 10^{-19}textrm{ J}
$$

Because the red light has the longest wavelength among visible light to get the work function it is enough the use the red light wavelength

$$
W=frac{hc}{lambda}=frac{1.24times 10^{-6}}{0.68times 10^{-6}}
$$

This gives that

$$
boxed{W=1.82textrm{ eV}}
$$

$$
W=1.82textrm{ eV}
$$

In order to solve this problem we should check up the recent data on how much solar energy hits the Earth per second. The recent NASA measurements state that $1360$J falls per second per square meter.

a) Now, we can calculate the amount that hits the square meter by simple multiplying of the rate by the sun hours

$$
I=1360times 3600times 3000=14.7times 10^9, frac{textrm{ J}}{textrm{m}^2}
$$

b) The final step is to find an area that will, with the given efficiency, provide the one year energy of an average US home.

$$
A=frac{E_{home}}{etimes I}=frac{4times 10^{11}}{0.2 times14.7times 10^{9}}=135textrm{ m}^2
$$

$$
textrm{a) }I=14.7times 10^9, frac{textrm{ J}}{textrm{m}^2}
$$

$$
textrm{b) }A=135textrm{ m}^2
$$

In order to solve a problem that involves finding of a wavelength of a particle we are going to use the de Broglie condition

$$
lambda=frac{h}{mv}
$$

Now, we can plug in the given values to have that required wavelength is

$$
lambda=frac{6.63times 10^{-34}}{9.11times 10^{-31}times 3times 10^6}
$$

$$
lambda=2.4times 10^{-10}textrm{ m}
$$

In order to solve this problem we are going to employ the de Broglie relation which says

$$
lambda=frac{h}{mv}
$$

from where the velocity can be expressed as

$$
v=frac{h}{mlambda}
$$

Now, we can plug in the given values to have that the speed of the electron is

$$
v=frac{6.63times 10^{-34}}{9.11times 10^{-31}times 3times 10^{-10}}
$$

$$
v=2.4times 10^{6}, frac{textrm{m}}{textrm{s}}
$$

a) We start with the conservation of energy in order to find the electron’s speed

$$
frac{mv^2}{2}=qV
$$

From which the speed is expressed as

$$
v=sqrt{frac{2qV}{m}}=sqrt{frac{2times 1.6times 10^{-19}times 5times 10^3 }{9.11times 10^{-31}}}
$$

Finally, we get that

$$
v=4.2times 10^{7}, frac{textrm{m}}{textrm{s}}
$$

b) Now, we use de Broglie relation to find the corresponding wavelength

$$
lambda=frac{h}{mv}=frac{6.63times 10^{-34}}{9.11times 10^{-31}times4.2times 10^7}
$$

$$
lambda=1.7times 10^{-11}textrm{ m}
$$

$$
v=4.2times 10^{7}, frac{textrm{m}}{textrm{s}}
$$

$$
lambda=1.7times 10^{-11}textrm{ m}
$$

We can do so by applying to formulas, the kinetic energy formula

$$
K=frac{mv^2}{2}
$$

and de Broglie relation

$$
lambda=frac{h}{mv}
$$

a) We start with the first relation and from it we can express the velocity as

$$
v=sqrt{frac{2K}{m}}
$$

After plugging the values

$$
v=sqrt{frac{2times0.025times 1.6times 10^{-19}}{1.67times 10^{-27}}}
$$

Finally

$$
v=2.2times 10^3,frac{textrm{m}}{textrm{s}}
$$

b) The de Broglie wavelength is then found directly from the second relation

$$
lambda=frac{h}{mv}
$$

After pluging the values

$$
lambda=frac{6.63times 10^{-34}}{1.67times 10^{-27}times2.2times 10^3}
$$

Finally

$$
lambda=1.8times 10^{-10}textrm{ m}
$$

$$
v=2.2times 10^3,frac{textrm{m}}{textrm{s}}
$$

$$
lambda=1.8times 10^{-10}textrm{ m}
$$

We can do so by applying to formulas, the kinetic energy formula

$$
K=frac{mv^2}{2}
$$

and de Broglie relation

$$
lambda=frac{h}{mv}
$$

and with the simple geometric formula

$$
C=2pr
$$

a) We start with the first relation and from it we can express the velocity as

$$
v=sqrt{frac{2K}{m}}
$$

After plugging the values

$$
v=sqrt{frac{2times 13.65times 1.6times 10^{-19}}{9.11times 10^{-31}}}
$$

Finally

$$
v=2.19times 10^6,frac{textrm{m}}{textrm{s}}
$$

b) The de Broglie wavelength is then found directly from the second relation

$$
lambda=frac{h}{mv}
$$

After pluging the values

$$
lambda=frac{6.63times 10^{-34}}{9.11times 10^{-31}times2.19times 10^6}
$$

Finally

$$
lambda=0.33times 10^{-9}textrm{ m}
$$

c) The circumference is found from the given formula and it is equal

$$
C=2pi r=2times 3.14times 0.519times 10^{-9}
$$

$$
C=3.26times 10^{-9} textrm{ m}
$$

Which is one order of magnitude more than the de Broglie wavelength.

$$
v=2.2times 10^6,frac{textrm{m}}{textrm{s}}
$$

$$
lambda=1.8times 10^{-10}textrm{ m}
$$

$$
C=3.26times 10^{-9} textrm{ m}
$$

In order to solve this problem we are going to start with the conservation of energy. The electrostatic potential energy that is used for acceleration is equal to the kinetic energy at the end of the process
$qV=frac{mv^2}{2}$
from where we can express the voltage as

$$
V=frac{mv^2}{2q}
$$

On the other hand, we can use the de Broglie relation to express the speed via the known variables

$$
lambda=frac{h}{mv}
$$

$$
v=frac{h}{lambda m}
$$

Now, we can insert the expression obtained for the speed into the expression for the voltage to get that

$$
V=frac{m}{2q}times frac{h^2}{lambda^2 m^2 }=frac{h^2}{2qmlambda^2}
$$

a) Let’s now calculate the required voltage in the case of an electron

$$
V=frac{6.63^2times 10^{-68}}{2times 1.6times 10^{-19}times 9.11times 10^{-31}times 0.18^2times 10^{-18}}
$$

Finally, we have that

$$
boxed{V=46.5textrm{ V}}
$$

b) The same thing, but with a different mass we do in the case of a proton

$$
V=frac{6.63^2times 10^{-68}}{2times 1.6times 10^{-19}times 1.67times 10^{-27}times 0.18^2times 10^{-18}}
$$

Finally, we have that

$$
boxed{V=0.025textrm{ V}}
$$

$$
textrm{a) }V=46.5textrm{ V}
$$

$$
textrm{b) }V=0.025textrm{ V}
$$

In order to solve this problem, we have to use the photoelectric effect formula which says that the maximum kinetic energy and the stopping potential are related as

$$
K_{max}=-qV_0
$$

Now, we can transfer our values from Joules to eVs but we can more elegantly note that eV is a unit charge times one Volt which gives that in our case

$$
K_{max}=-(-1e)times 3.8V
$$

Finally

$$
boxed{K_{max}=3.8textrm{ eV}}
$$

$$
K_{max}=3.8textrm{ eV}
$$

In order to solve this problem we are going to use the fact that the work function is given as

$$
W=hf_0
$$

Now, we can plug in the values in the above expression to have that

$$
W=6.63times 10^{-34}times 8times 10^{14}
$$

Finally we have that the work function is given

$$
boxed{W=5.3 times 10^{-19}textrm{ J}}
$$

$$
W=5.3 times 10^{-19}textrm{ J}
$$

This problem involves a metal of a known threshold frequency and the light of a known frequency.
In order to solve this problem, we are going to use the photoelectric effect equation which tells us that the maximum kinetic energy is given as

$$
K_{max}=E_{ph}-W=hf-hf_0
$$

Now, we can plug in the given values which gives us that the maximum kinetic energy is

$$
K_{max}=h(f-f_0)=6.63times 10^{-34}times(1.6times 10^{15}-8times 10^{14})
$$

Finally, we have that

$$
boxed{K_{max}=5.3 times 10^{-19}textrm{ J}}
$$

$$
K_{max}=5.3 times 10^{-19}textrm{ J}
$$

We can do so by using de Broglie relation which is given as follows

$$
lambda=frac{h}{mv}
$$

Now, we can plug in the values to obtain that de Broglie wavelength is

$$
lambda=frac{6.63times 10^{-34}}{3.3times 10^{-27}times 2.5times 10^4}
$$

Finally,

$$
boxed{lambda=8times 10^{12}textrm{ m}}
$$

$$
lambda=8times 10^{12}textrm{ m}
$$

To do so, we are going to use the work function equation which tells us that
$W=frac{hc}{lambda_0}$
and photoelectric effect equation

$$
K=E_{ph}-W
$$

a) Now, let’s find the the threshold wavelength from the first equation. We can express it as

$$
lambda_0=frac{hc}{W}=frac{1240times 10^{-9}}{4.7}
$$

Finally

$$
boxed{lambda=263times 10^{-9}textrm{ m}}
$$

b) The final step is to find the maximum kinetic energy from the second equation

$$
K_{max}=E_{ph}-W=frac{hc}{lambda}-W=frac{1240times 10^{-9}}{150times 10^{-9}}-4.7
$$

Which gives

$$
boxed{K_{max}=3.57textrm{ eV}}
$$

$$
lambda=263times 10^{-9}textrm{ m}
$$

$$
K_{max}=3.57textrm{ eV}
$$

This can be done by employing the work function equation which tells us that

$$
W=frac{hc}{lambda_0}
$$

From where we can express the threshold wavelength as

$$
lambda_0=frac{hc}{W}
$$

Now, we can plug in the given values to get that threshold wavelength is

$$
lambda_0=frac{1240times 10^{-9}}{2.48}
$$

Finally, we have

$$
boxed{lambda_0=500times 10^{-9}textrm{ m}}
$$

$$
lambda_0=500times 10^{-9}textrm{ m}
$$

We can find the value of the electron’s speed mentioned above by using de Broglie relation

$$
lambda=frac{h}{mv}
$$

a) We can take now the first equation and express the speed from it as follows

$$
v=frac{h}{mlambda}=frac{6.63times 10^{-34}}{9.11times 10^{-31}times 400times 10^{-9}}
$$

Finally, we have that

$$
boxed{v=1.82times 10^3, frac{textrm{m}}{textrm{s}}}
$$

On the other hand we will use the standard kinetic energy equation to find electron’s kinetic energy.

$$
K=frac{mv^2}{2}
$$

b) Now, by simply inserting the values into the second equation and dividing by the energy of one eV in Joules we obtain that

$$
K=frac{9.11times 10^{-31}times 1.82^2times 10^6}{2times 1.6times 10^{-19}}
$$

$$
boxed{K=9.43times 10^{-6}{textrm { eV}}}
$$

$$
textrm{a) }v=1.82times 10^3, frac{textrm{m}}{textrm{s}}
$$

$$
textrm{b) }K=9.43times 10^{-6}{textrm{ eV}}
$$

We can do so, by first expressing the electron’s speed from its de Broglie wavelength and then inserting it into the kinetic energy expression in the manner as follows

$$
lambda=frac{h}{mv}
$$

so the speed is

$$
v=frac{h}{mlambda}
$$

Now, the kinetic energy is given as

$$
K=frac{mv^2}{2}=frac{m}{2}times frac{h^2}{m^2lambda^2}
$$

Now, we can plug in the values into the above’s expression to get that

$$
K=frac{h^2}{2mlambda^2times 1.6times 10^{-19}}=frac{6.63^2times 10^{-68}}{2times 9.11times 10^{-31}times 20^2times 10^{-18}times 1.6times 10^{-19}}
$$

Finally, we have that

$$
boxed{K=3.8times 10^{-3}textrm{ eV}}
$$

$$
K=3.8times 10^{-3}textrm{ eV}
$$

a) To find the threshold wavelength, we are going to use the frequency-wavelength relation since the former is given in the problem. We have that

$$
c=lambda f
$$

Now, we can use this equation to express the threshold wavelength and after we substitute the given values we have that

$$
lambda=frac{c}{f}=frac{3times 10^{8}}{1.2times 10^{15}}
$$

Which gives that

$$
boxed{lambda=250times 10^{-9}textrm{ m}}
$$

b) The next step is finding the work function of the tin which is by definition given as

$$
W=hf_0
$$

If we now take the above equation and after we plug in the values we obtain that the work function is

$$
W=6.63times 10^{-34}times 1.2times 10^{15}
$$

that gives us

$$
W=8times 10^{-19}textrm{ J}
$$

or in eV

$$
W=frac{8times 10^{-19}}{1.6times 10^{-19}}=boxed{5textrm{ eV}}
$$

a) To find the maximum kinetic energy of the ejected electrons we are going to use the photoelectric effect relation which is given as

$$
K_{max}=frac{hc}{lambda}-W
$$

where $hc=1240times 10^{-9}$ eV$cdot$m.

Now, we can plug in the given values to have that the maximum kinetic energy of the ejected electrons is

$$
K_{max}=frac{1240times 10^{-9}}{167times 10^{-9}}-5=7.43-5
$$

Finally, we have that

$$
boxed{K_{max}=2.43textrm{ eV}}
$$

$$
lambda=250times 10^{-9}textrm{ m}
$$

$$
W=5textrm{ eV}
$$

$$
K_{max}=2.43textrm{ eV}
$$

a) In order to find the energy of a single photon we are going to use the photon energy equation which tells us that

$$
E_{ph}=frac{hc}{lambda}
$$

Now, we can simply plug in the given value of the wavelength in the above equation to obtain that

$$
E_{ph}=frac{6.63times 10^{-34}times 3times 10^{8}}{632.8times 10^{-9}}=boxed{3.14times 10^{-19}textrm{ J}}
$$

b) To find the number of photons with this energy that are emitted by the laser each second we are going to use the fact that the amount of energy emitted per second is defined as power so we have that

$$
dot n=frac{P}{E_{ph}}
$$

Now we can simply plug in the given value of the power and the value obtained for the single photon energy to have that

$$
dot n=frac{5times 10^{-4}}{3.14times 10^{-19}}=boxed{15.9times 10^{14}, frac{textrm{photons}}{textrm{s}}}
$$

$$
textrm{a) }E_{ph}=3.14times 10^{-19}textrm{ J}
$$

$$
textrm{b) }dot n=15.9times 10^{14}, frac{textrm{photons}}{textrm{s}}
$$

Firstly, we should understand that the power of the light hitting a surface equals the product of the intensity and the area of the surface

$$
P=IA=Ipi r^2=1.5times 10^{-11}times 3.14times 3.5^2times 10^{-6}
$$

$$
boxed{P=5.8times 10^{-16}textrm{ W}}
$$

a) In order to find the number of photons that each second hit the pupil we need an energy of a single photon so we are going to use the photon energy equation which tells us that

$$
E_{ph}=frac{hc}{lambda}
$$

Now, we can simply plug in the given value of the wavelength in the above equation to obtain that

$$
E_{ph}=frac{6.63times 10^{-34}times 3times 10^{8}}{550times 10^{-9}}=boxed{3.6times 10^{-19}textrm{ J}}
$$

b) Finally, we can find the number of photons that are received by the eye each second by using the fact that the amount of energy emitted per second is defined as power so we have that

$$
dot n=frac{P}{E_{ph}}
$$

Again we can simply plug in the given value of the power and the value obtained for the single photon energy to have that

$$
dot n=frac{5.8times 10^{-16}}{3.6times 10^{-19}}=boxed{1.6times 10^{3}, frac{textrm{photons}}{textrm{s}}}
$$

$$
textrm{a) }E_{ph}=3.6times 10^{-19}textrm{ J}
$$

$$
textrm{b) }dot n=1.6times 10^{3}, frac{textrm{photons}}{textrm{s}}
$$

In this problem we should make a plot from the given data, stopping potential vs frequency but initially we have to convert wavelength to the frequency. Then we should check the plot to find the intercept line, the work function, the slope and $h/q$ value which should be compared to the accepted value.

First, let’s convert the wavelengths given in the table to frequencies by using the formula that connects the two parameters

$$
f=clambda
$$

begin{table}[]
begin{tabular}{|r|r|r|}
hline
multicolumn{1}{|l|}{$lambda$(nm)} & multicolumn{1}{l|}{f(Hz)} & multicolumn{1}{l|}{$V_0$(eV)} \ hline
200 & $15times 10^{14}$ & 4.2 \ hline
300 & $10times 10^{14}$ & 2.06 \ hline
400 & $7.5times 10^{14}$ & 1.05 \ hline
500 & $6times 10^{14}$ & 0.41 \ hline
600 & $5times 10^{14}$ & 0.03 \ hline
end{tabular}
end{table}

Now, we can calculate the slope of the function by simply taking any two points and dividing their displacements with each other

$$
Delta V_0=kDelta f
$$

$$
k=frac{Delta V_0}{Delta f}
$$

We can take the last point and the first point that to have

$$
k=frac{4.17}{1times 10^{15}}=4.17times 10^{-15}, frac{textrm{ V}}{textrm{ Hz}}
$$

and since 1V=1J/1C we can write

$$
boxed{k=4.17times 10^{-15}frac{textrm{ J}}{textrm{C}cdottextrm{ Hz}}}
$$

On the other hand the accepted value is set to be

$$
frac{h}{e}=frac{6.63times 10^{-34}}{1.6times 10^{-19}}=boxed{4.14times 10^{-15}frac{textrm{ J}}{textrm{C}cdottextrm{ Hz}}}
$$

we see that the difference is relatively small and equals $approx 10^{-17}$.

From the graph we read that the threshold frequency, (the interception point) is approximately given as $f_0=4.99times 10^{14}$. Now, the threshold wavelength can be found as

$$
lambda_0=frac{c}{f_0}=frac{3times 10^8}{4.99times 10^{15}}
$$

$$
boxed{lambda_0=601times 10^{9}textrm{ m}}
$$

Finally, the work function can be found from the following well-known formula

$$
W=hf_0=6.63times 10^{-34}times 4.99times 10^{14}
$$

$$
boxed{W=3.3times 10^{-19}textrm{ J}}
$$

$$
k=4.17times 10^{-15}frac{textrm{ J}}{textrm{C}cdottextrm{ Hz}}
$$

$$
lambda_0=601times 10^{9}textrm{ m}
$$

$$
W=3.3times 10^{-19}textrm{ J}
$$

We can now use the well known spring equation which gives us the spring force

$$
F=kx
$$

from where we can express the spring constant $k$ as

$$
k=frac{F}{x}=frac{400}{0.15}=boxed{2.7times 10^3, frac{textrm{N}}{textrm{m}}}
$$

$$
k=2.7times 10^3, frac{textrm{N}}{textrm{m}}
$$

In order to solve this problem we are going to apply the Coulomb law which states that the force between two charges is

$$
F=frac{1}{4pi varepsilon_0}cdotfrac{q_1q_2}{r^2}
$$

From where we can express the second charge as

$$
q_2=frac{4pivarepsilon_0cdot Fcdot r^2}{q_1}
$$

Now, we can plug in the values in the expression obtained above to have that

$$
q_2=frac{9times 0.02^2}{9times 10^9times 8times 10^{-7}}
$$

Finally, we have that

$$
boxed{q_2=5times 10^{-7}textrm{ C}}
$$

$$
q_2=5times 10^{-7}textrm{ C}
$$

For this purpose we are going to use Ohm’s law which gives us the relation between the voltage and the resistance of the circuit

$$
I=frac{V}{R}
$$

Now, we have that the total current is equal to twelve times current in the single light set

$$
I=12I_s=12frac{V}{R_s}=12timesfrac{120}{24times 6}
$$

Which gives that the total current is

$$
boxed{I=10 textrm{ A}}
$$

$$
I=10 textrm{ A}
$$

We can solve this problem by employing the well known equation that defines the force on a current-carrying wire in a magnetic field

$$
F=ILB
$$

From this expression we can express the current in a single step as follows

$$
I=frac{F}{LB}=frac{1.1times 10^{-3}}{1.2times 5times 10^{-5}}
$$

Finally we have that

$$
boxed{I=18.3textrm{ A}}
$$

$$
I=18.3textrm{ A}
$$