Work function of sodium and copper is 2.3 eV and 4.5 eV respectively.Ratio of their threshold wavelength will be…..

Options

A. `1:2`
B. `2:1`
C. `4:1`
D. `1:4`

Answer

B

Explanation

In `phi=hf_(0)=(h_(c))/(lambda_(0)),lambda_(c)`=constant
`therefore phi prop (1)/(lambda_(0))`
`(phi_(Na))/(phi_(cu))=(lambda_(0)(cu))/(lambda_(0)(Na))`
`(2.3)/(4.5)=(lambda_(0)(cu))/(lambda_(0)(Na))`
`(1)/(2)~~(lambda_(0)(cu))/(lambda_(0)(Na)) therefore (lambda_(0)(cu))/(lambda_(0)(Na))=(2)/(1)`