A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are `u_1` and `u_2` respectively. If the ratio `u_1: u_2 = 2:1 and hc= 1240 eV` nm, the work function of the metal is neraly. (a)3.7 eV (b) 3.2 eV (c ) 2.8eV (d) 2.5eV.

Options

A. `3.7 eV`
B. `3.2 eV`
C. `2.8 eV`
D. `2.5 eV`

Answer

A

Explanation

`K_("max")=(hc)/lambda - phi`
`1/2 m u_(0)^(2)=1240/248- phi` …(1)
`1/2 m u_(2)^(2)=1240/310 - phi` …(2)
Dividing, the two equations
`(u_(1)/u_(2))^(2)=(5-phi)/(4-phi)` & `u_(1)/u_(2)=2/1` (given)
`4=(5-phi)/(4-phi) implies phi =11/3=3.7 eV`