Question

A 3.00-W beam of light of wavelength 122 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.
Part A
What is the work function (in electron volts) of this metal?
ϕ = ??? eV

Part B
How many photoelectrons are ejected each second from this metal?
N = ??? electrons/s

Part C
If the power of the light beam, but not its wavelength, were reduced by half, what would be the answer to part B?
N = ??? electrons/s

Part D
If the wavelength of the beam, but not its power, were reduced by half, what would be the answer to part B?
N = ??? electrons/s

Answer 1

The maximum kinetic energy of ejected electrons from the metal is expressed as

where, Emax is the maximum kinetic energy of ejected electron, vmax is the maximum velocity, W0 is the work function of the metal. Plank Constant = h = 6.626 × 10-34 m2 kg/s, Velocity of light in free space = c = 2.997 x 108 m/s, 1 eV = 1.602 x 10-19 J.

Part - A:

The work function of the metal will be

Part - B:

The energy of the photoelectron is

The beam power is P = 3.00 W.

Photoelectrons are ejected each second from this metal will be

Part - C:

The number of ejected electrons will be half if the incident power is halved, provided there is no change in the wavelength. Therefore, in this case, the value of N will be

Part - D:

If the wavelength of the beam, but not its power, were reduced by half, the kinetic energy of the ejected electron will be

Hence, photoelectrons are ejected each second from this metal will be