Question

A 2.00-W beam of light of wavelength 126 nm falls on a metal surface. You observe that the maximum kinetic energy of the ejected electrons is 4.20 eV. Assume that each photon in the beam ejects a photoelectron.
(a) What is the work function (in electron volts) of this metal?
(b) How many photoelectrons are ejected each second from this metal?
(c) If the power of the light beam, but not its wavelength, were reduced by half what would be the answer to part (b)?
(d) If the wavelength of the beam, but nots its power, were reduced by half what would be the answer to part (b)?

Answer 1

a) 5.67 eV

b) $1.27\cdot 10^{18}$

c) $6.33\cdot 10^{17}$

d) $6.33\cdot 10^{17}$

Explanation:

a)

The photoelectric effect occurs when light is shone on the surface of the metal, and electrons are released from the surface of the metal if the incoming photons have enough energy.

The equation that describes the photoelectric effect is

$\frac{hc}{\lambda}=\phi+K_{max}$ (1)

where

the first term on the left is the energy of the incoming photons, where

h is the Planck constant

c is the speed of light

$\lambda=126 nm=126\cdot 10^{-9}m$ is the wavelength of the incoming photons

$\phi$ is the work function of the metal

$K_{max}$ is the maximum kinetic energy of the photoelectrons

The energy of the photons is (in eV)

$\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{126\cdot 10^{-9}}=1.58\cdot 10^{-18} J (\cdot \frac{1}{1.6\cdot 10^{-19}J/eV})=9.87 eV$

So, the work function of this metal is:

$\phi=\frac{hc}{\lambda}-K_{max}=9.87-4.20=5.67 eV$

b)

The power of the beam of light is

P = 2.00 W

Which means that every second, the energy emitted is 2.00 J; in electronvolts,

$E=2.00 J \cdot \frac{1}{1.6\cdot 10^{-19}}=1.25\cdot 10^{19}eV$

Each photon has a energy of (calculated in part a)

$E_1=9.87 eV$

Therefore, the number of incoming photons per second is

$N=\frac{E}{E_1}=\frac{1.25\cdot 10^{19}}{9.87}=1.27\cdot 10^{18}$

In the photoelectric effect, the ratio photons:photoelectrons is 1:1, because each incoming photon "hits" only one electron, giving energy to it. Therefore, the number of photoelectrons emitted is equal:

$N=1.27\cdot 10^{18}$

c)

In this case, the power of the light beam is halved; so, the new power is

P = 1.00 W

This means that the energy emitted per second is now

$E=1.00 J$

In electronvolts,

$E=1.00\cdot \frac{1}{1.6\cdot 10^{-19}}=6.25\cdot 10^{18} eV$

And therefore, the number of incoming photons is:

$N=\frac{E}{E_1}=\frac{6.25\cdot 10^{18}}{9.87}=6.33\cdot 10^{17}$

And therefore, since the ratio photons:photoelectrons is 1:1, the number of photoelectrons emitted is:

$N=6.33\cdot 10^{17}$

d)

In this case, the wavelength of the beam is reduced by half, so the new wavelength is

$\lambda=\frac{126}{2}=63 nm$

Looking at eq.(1), the equation of the photoelectric effect, we see that this change will affect the energy of the incoming photons. In particular, the new energy of the photons will be:

$\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{63\cdot 10^{-9}}=3.16\cdot 10^{-18} J \cdot (\frac{1}{1.6\cdot 10^{-19}J/eV})=19.7 eV$

This means that the energy of the photons is now twice the previous energy: this also means that the emitted photoelectrons will have a larger maximum kinetic energy (since more energy is given off by the photons).

Since the power is still 2.00 W, the total energy emitted per second is $1.25\cdot 10^{19}eV$ (part b), therefore the number of photons emitted per second is:

$N=\frac{E}{E_1}=\frac{1.25\cdot 10^{19}}{19.7}=6.33\cdot 10^{17}$

So, this is also the number of photoelectrons emitted.