# Solutions 10

- Page ID
- 47386

## Q10.1

What is the wavelength associated with a photon of a light with the energy is 3.6x10^{-19}J?

## S10.1

$$E=h\nu$$

E is the energy of photon, h is the planck's constant, \(\nu\) is the frequency of the light. Therefore,

$$\nu=\dfrac{E}{h}=\dfrac{3.6x10^{-19} J}{6.63x10^{-34}m^{2}kg s^{-1}}=5.4 x 10^{14} s^{-1}$$

$$c=\lambda\nu$$

c is the speed of light, \(\lambda\) is wavelength.

$$\lambda=\dfrac{c}{\nu}=\dfrac{3.0x10^{8} m s^{-1}}{5.4 x 10^{14} s^{-1}}=5.6 x 10^{-7} m=560 nm$$

## Q10.2

Calculate the energy of a photon of a light with the frequency 6.5x10^{-14} s^{-1} ?

## S10.2

$$E=h\nu=(6.63 x 10^{-34} m^{2} kg s^{-1}) x (6.5 x 10^{-14} s^{-1})=4.3 x 10^{-47} J$$

## Q10.3

Rigel is the brightest star in the constellation Orion and the seventh brightest star in the night sky. Rigel has an emission spectrum that peaks at ~145 nm. What is the surface temperature of Rigel? How does this temperatures compare with our sun (you have to look this up)?

## S10.3

According to Wien's displacement law, which you can derive from the blackbody radiation formula from the lecture notes (for details about the derivation, refer to wikipedia: https://en.wikipedia.org/wiki/Wien%2...splacement_law), we have:

$$\lambda_{max}=\dfrac{2.8977729x10^{-3} m K}{T}$$

\(\lambda_{max}\) stands for the wavelength that corresponds to the maximum/peak of the blackbody emission spectrum at a certain temperature T.

So,

$$T=\dfrac{2.8977729x10^{-3} m K}{\lambda_{max}}=\dfrac{2.8977729x10^{-3} m K}{145 x 10 ^{-9} m}=2.0 x 10^{4} K$$

The temperature of the sun is 5778 K. The temperature of Rigel is about 3.5 times that of the sun.

## Q10.4

Find the longest-wavelength photon that can eject an electron from potassium, given that the work function is 2.24 eV. Is this visible electromagnetic radiation?

## S10.4

In photoelectric effect, the maximum kinetic energy of the ejected electron is:

$$KE_{e}=h\nu-\Phi$$

\(KE_{e}\) is the maximum kinetic energy of the ejected electron;

\(h\nu\) is the photon's energy;

\(\Phi\) is the work function.

Since \(E=h\nu=\dfrac{hc}{\lambda}\), the longer the wavelenght of the photon, the less energy it has.

Therefore, the longest wavelength photon that can eject an electron will lead to almost zero kinetic energy for the ejected electron:

$$0=h\nu-\Phi$$

$$\Phi=h\nu=\dfrac{hc}{\lambda}$$

$$\lambda=\dfrac{hc}{\Phi}=\dfrac{(6.63 x 10^{-34} m^2 kg s^{-1}) x (3.0 x 10^8 m s^{-1})}{2.24 eV x \dfrac{1.6 x 10^{-19} J}{1 eV}}=5.55 x 10^{-7} m = 555 nm$$

Visible E&M radiation is in the range of 400 nm - 700 nm. So this wavelength is in the visible range.

## Q10.5

A laser with a power output of 2.00 mW at a wavelength of 400 nm is projected onto calcium metal. (a) How many electrons per second are ejected? (b) What power is carried away by the electrons, given that the work function is 2.71 eV?

## S10.5

(a) The energy for a single photon of 400 nm wavelength is:

$$E=h\nu=\dfrac{hc}{\lambda}=\dfrac{(6.63 x 10^{-34} m^2 kg s^{-1}) x (3.0 x 10^8 m s^{-1})}{400 x 10^{-9} m}=4.97 x 10^{-19} J$$

Power output of 2.00 mW means there are 2.00 mJ of energy deposited on the calcium metal per second through absorption of 400 nm photons. 2.00 mJ of energy corresponds to \(\dfrac{2 x 10^{-3} J}{4.97 x 10^{-19} J}=4.02 x 10^{15}\) photons.

Each photon will eject one electron, therefore, \(4.02x10^{15}\) electrons will be ejected per second.

(b) The energy that is carried away by each electron is the kinetic energy that it has after being ejected:

$$KE_{e}=h\nu-\Phi=4.97x10^{-19} J - 2.71 eVx \dfrac{1.60 x 10^{-19} J}{1 eV}=6.34 x 10^{-20} J$$

The power (\(P_{e}\)) carried away by the electrons is the energy that is carried away by all the ejected electrons per second:

$$P_{e}=(4.02x10^15 s^{-1}) x (6.34 x 10^{-20} J)=0.255 mW$$

## Q10.6

What is the longest-wavelength electromagnetic radiation that can eject a photoelectron from silver, given that the work function is 4.73 eV? Is this in the visible range?

## S10.6

Following the same argument as Q10.4, we have:

$$\lambda=\dfrac{hc}{\Phi}=\dfrac{(6.63 x 10^{-34} m^2 kg s^{-1}) x (3.0 x 10^8 m s^{-1})}{4.73 eV x \dfrac{1.6 x 10^{-19} J}{1 eV}}=2.62 x 10^{-7} m = 262 nm$$

Visible E&M radiation is in the range of 400 nm - 700 nm. So this wavelength is not in the visible range.