Two photons of energy 2.5 eV and 3.5 eV fall on a metal surface on work function 1.5 eV. The ratio of the maxium velocities of the photoelectrons emitted from the metal surface is

Options

A. `1:4`
B. `2:1`
C. `1:2`
D. `1:sqrt2`

Answer

D

Explanation

`(1)/(2)mv(2)/(1)=hv_(a)-phi_(o)`
and`(1)/(2)mv(2)/(2)=hv_(a)-phi_(o)`
Divinding equation (1) and (2),
`((1)/(2)mv(2)/(1))/((1)/(2)mv(2)/(2))=(hv_(a)-phi_(o))/(hv_(b)-phi_(o)implies (v_(1)2)/(v_(2)^(2)))=((2.5-1.5))/((3.5-1.5))`
`implies v_(1):v_(2)=1:sqrt2`