Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively, successively illuminate a metallic surface whose work function is 0.5 eV. Ratio of maximum speeds of emitted electrons will be

Options

A. `1:4`
B. `1:1`
C. `1:5`
D. `1:2`

Answer

D

Explanation

`hv=w+1//2mv_("max")^(2)`
`i.e. E=w+1//2v_("max")^(2)`
`:.1//2mv_("max")^(2)=E-w`
`:.(v_(1)^(2))/(v_(2)^(2))=(E_(1)-w)/(E_(2)-w)`
`(1eV-0.5eV)/(2.5eV-0.5eV)`
`=(0.5)/(2)=(1)/(4)`
`:.(v_(1))/(v_(2))=1//2`