Question

Light of wavelength 400 nm is incident on a metal surface. The stopping
potential for the resulting electrons is 0.91 V.
a) What is the work function of the metal?
b) Identify the metal. (Use the Table)
c) What is the cut-off frequency for this target?

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Answer 1

Here we us Einstein's photoelectric equation as,

* = hv, +eV

  

== 0 + eV

h = plancks constant = 6.63 x 10-34Js

a = Incident light wavelength = 400 nm = 400 x 10-ºm

Vo = cut - off frequency =?

Vo = stopping potential = 0.91 V

e = magnitude of charge on electron = 1.6 x 10-19 C

C = spped of light = 3 X 108 m/s

0o = workfunction of metal = hv.

a) Let us first find work function,

hc Do =-ev

  

Φο =- 6.63 Χ 10-34x3x 108 -- (1.6 x 10-19 x 0.91) 400 x 10-9

  

Φ, = 3.517 Χ 10-19

1 eV = 1.6 x 10-19)

· Work function in terms of eV can be,

3.517 X 10-19 1.6 x 10-19 a = 2.19 eV 2.2 eV

This is the work function of the metal

b) from the calculation of work function given metal is potassium

c) cut-off frequency (v.) can be determined by using the formula of workfunction as,

Φο = hυο

Vo = = 3.517 x 10-19 6.63 x 10-34 = 5.31 x 1015 Hz h