Question

A beam of UV light with wavelength of 200 nm is incident on a metal. If the stopping potential of the photoelectrons is 2.8 V, determine the work function and cut-off frequency of the metal.

Answer 1

The work function of the metal is approximately 1.17 x 10^-19 J or 3.07 eV, and the cut-off frequency is approximately 1.5 x 10^15 Hz.

Step-by-step explanation:

The work function can be determined using the equation:

hf = Φ + Ek(max)

where h is Planck's constant (6.626 x 10^-34 J s), f is the frequency of the incident light, Φ is the work function, and Ek(max) is the maximum kinetic energy of the emitted electrons.

The cut-off frequency is given by the equation:

f(cut-off) = v(speed of light) / λ(wavelength)

Substituting the given values, we have:

Φ = hf - Ek(max) = (6.626 x 10^-34 J s)(3 x 10^8 m/s) / (200 x 10^-9 m) - 2.8 eV

and

f(cut-off) = (3 x 10^8 m/s) / (200 x 10^-9 m)

Therefore, the work function of the metal is approximately 1.17 x 10^-19 J or 3.07 eV, and the cut-off frequency is approximately 1.5 x 10^15 Hz.