Question

light of wavelength 450nm is incident on on metal target (work function 1.8ev) in a phototube what is the valu of stopping potential?

Answer 1

Light of wavelength 450 nm incident on the metal surface

So energy of light that incident on the surface is given by

$E = \frac{hc}{\lambda}$

plug in all values in this equation

$E = \frac{6.6 * 10^{-34}* 3* 10^8}{450*10^{-9}}$

$E = 4.4 * 10^{-19} J$

now energy in eV units is given by

$E = \frac{4.4*10^{-19}}{1.6*10^{-19}}$

$E = 2.75 eV$

now by Einstein's Equation of photoelectric effect we can say

$E = W + eVo$

here

Vo = stopping potential

W = work function

now plug in above values in this equation

2.75 eV = 1.8 eV + eVo

eVo = (2.75 - 1.8)eV

eVo = 0.95 eV

Vo = 0.95 Volts

so the stopping potential will be 0.95 Volts