Mercury violet `(lambda = 4558 Å)` is falling on a photosensitive material `(phi = 2.5 eV)`. The speed of the ejected electrons is in `ms^(-1)` , about

Options

A. `3 xx 10^(5)`
B. `2.65 xx 10^(5)`
C. `4 xx 10^(4)`
D. `3.65 xx 10^(7)`

Answer

B

Explanation

By Using `E = W_(0) + (1)/(2) mv_(max)^(2)` , where
`E = (12375)/(4558) = 2.71 eV`
`rArr 2.71 ev = 2.5 eV + (1)/(2) eV + (1)/(2) xx 9.1 xx 10^(-31) xx v_(max)^(2) `
`rArr 0.21 xx 1.6 xx 10^(-19) = (1)/(2) xx 9.1 xx 10^(-31) xx v_(max)^(2)`
`rArr v_(max) = 2.65 xx 10^(5) m//s`