Question

when a metallic surface is illuminated with radiation of wavelength λ, the stopping potential is V if the same surface is illuminated with radiation of wavelength 2λ,the stopping potential is V/4.the threshold wavelength for the metallic surface is ​

Answer 1

Q]_____?

=> 4 λ

Explanation:

Radius in magnetic field of circular orbit,

R = mV / qB = (√2mE) /qB

and Total energy of moving particle in a circular orbit, E = ( q^2 B^2 R^2 ) / 2m

For a Proton enter in a region of magnetic field,

E1 = ( e^2 × B^2 × R^2 ) / 2 × mp _[i]

where M is the mass of proton

Similarly,for a α-particle moves in a uniform magnetic field

E2 =[( 2e )^2 × B^2 × R^2] / 2 × (4mp)

[ ∵ mα = 4mp ] _[ii]

Dividing Eq [ii] by Eq [ii] we get,

E2/E1 = { [ (2e)^2 × B^2 × R^2 / 2 × (4mp) ] × [ 2 × mp / e^2 × B^2 × R^2] }

E2/E1 = 1 => E2 = E1 = 1MeV

The threshold wavelength for the metallic surface is 4 λ !