Question

The photoelectric work function for aluminium is 4.2 ev .What is the stopping potential for radiation of wavelength 2500a°

Answer 1

The stopping potential for the wavelength 2500 A° = 0.45 volts

Given , the stopping potential(Φ) of aluminum = 4.2 eV

= 4.2 × 1.6 × (10)^-19 J

= 7.2 × (10)^-19 J

Wavelength of radiation = λ = 2500A°

= 2500 × (10)^-10 m

Energy associated with the wavelength = hc/λ

c - speed of light

h - Planck's constant

Energy = (6.6 × (10)^-34 × 3 × (10)^8)/(25 × (10)^-8)

Energy = 7.92 × (10)^-19 J

Let electric potential be V.

Using Einstein's photoelectric equation to find the stopping potential -

eV = hc/λ - Φ

e - change of an electron

=> eV = 7.92 × (10)^-19 - 7.2 × (10)^-19

=> eV = 0.72 × (10)^-19

=> V = 0.72 × (10)^-19/1.6 × (10)^-19

=> V = 0.45 volts

The stopping potential for the wavelength 2500 A° = 0.45 volts