When a metal surface is illuminated by light wavelengths `400 nm` and `250 nm` , the maximum velocities of the photoelectrons ejected are `upsilon` and `2 v` respectively . The work function of the metal is
`( h = "Planck's constant" , c = "velocity of light in air" )`

Options

A. `2hc xx 10^6 J`
B. `1.5 hc xx 10^6 J`
C. `hc xx 10^6 J`
D. `0.5 hc xx 10^6 J`

Answer

A

Explanation

By using `(hc)/(lamda) = W_0 + 1/2 mv^2`
` rArr (hc)/(400 xx 10^(-9) ) = W_0 + 1/2 mv^2` ….(i)
and `(hc)/(250 xx 10^(-9)) = W_0 + 1/2 m (2v)^2` ….(ii)
On solving (i) and (ii)
`1/2mv^2 = (hc)/(3) [ (1)/(250 xx 10^(-9) ) - (1)/(400xx 10^(-9) ) ]` ....(iii)
from equation (i) and (iii) `W_0 = 2hc xx 10^6 J`