The K.E. of the most energetic electrons emitted from a metallic surface is doubled when the wavelength, `lambda` of the incident radiation is reduced from 400 nm to 310 nm. The work function of the metal is

Options

A. 0.9 eV
B. 2.2 eV
C. 1.7 eV
D. 3.1 eV

Answer

B

Explanation

`(1)/(2)mv^(2)=(hc)/(lambda)-phi`
and `2xx(1)/(2)mv^(2)=(hc)/(lambda)-phi`
Dividing (ii) by,(i) we get `2=(hc//lambda.-phi)/(hc//lambda-phi)`
or `(2hc)/(lambda)-2phi=(hc)/(lambda.)-phi`
`or phi=(2hc)/(lambda)-(hc)/(lambda.)`
`=(2xx6.63xx10^(-34)xx3xx10^(8))/(4xx10^(-7)xx1.6xx10^(-19))=2.2eV`