Question

A 620 nm light falls on a photoelectric surface and electrons with the maximum kinetic energy of 0.14 eV are emitted. (a) Determine the work function (in eV). eV (b) Determine the cutoff frequency of the surface (in THz). THz (c) What is the stopping potential (in V) when the surface is illuminated with light of wavelength 420 nm

Answer 1

Answer:

(a) The work function is 1.86 eV.

(b) The cut off frequency is 450 THz.

(c) The stopping potential is 1.16 V.

Explanation:

incident wavelength = 620 nm

Kinetic energy, K = 0.14 eV

According to the photoelectric equation

E = W + KE

where, W is the work function, KE is the kinetic energy.

(a) Let the work function is W.

$W = E - KE\\W = \frac{h c}{\lambda }- KE\\W =\frac{6.63\times 10^{-34}\times3\times 10^{8}}{620\times 10^{-9}\times 1.6\times 10^{-19}}-0.14\\\\W =1.86 eV$

(b) Let the cut off frequency is f.

W = h f

$1.86\times 1.6\times 10^{-19} = 6.63\times 10^{-34}\times f\\f = 4.5\times 10^{14} Hz =450 THz$

(c) Let the stopping potential is V.

$E = W + eV\\\frac{6.63\times 10^{-34}\times 3\times 10^{8}}{420\times 10^{-9}\times 1.6\times 10^{-19}}=1.8 + eV\\\\V = 1.16 V$