When a metal surface is illuminated by a light of wavelength 400nm and 250 nm . The maximum velocites of the photo electrons ejected are v and 2v respectively. The work function of the metal is (h= Planck's constant e= Velocity of light in air

Options

A. `2hexx10^(6)J`
B. `1.5hexx10^(6)J`
C. `hcxx10^(6)J`
D. `0.5hcxx10^(6)`

Answer

A

Explanation

Given that ,
Wavelength ,`lamda_(1)`=400nm
Wavelength ,`lamda_(2)`250nm
Velocity ,`v_(1)`=v
Velocity,`V_(2)`=2v
The expression for Einstein's Photoelectric equation given as
`(1)/(2)mv_max^2=(hc)/(lamda)-w`
For ,`V_(1)andlamda_(1)`we get ,
`(1)/(2)m(v_(1))_(max)^(2)=(hc)/(lamda_(1))-w`
For ,`V_(2)and lamda_(2)`,we get ,
`(1)/(2)m(v_(2))_max^(2)=(hc)/lamda_(2)-w`
`therefore((1)/(2)m(v_(1))_(max)^(2))/((1)/(2)m(v_(2))_max^(2))=((hc)/(lamda_(1))-w)/((hc)/lamda_(2)-w)`
`((v_(1)^(2))max)/((v_(2))_max^(2))=((hc)/(lamda_(1))-w)/((hc)/lamda_(2)-w)`
`Implies (v_(2))/((2_v)^(2))=((hc)/(lamda_(1))-w)/((hc)/lamda_(2)-w)`
`implies (hc)/(lamda2)-w=4(hc)/(lamda_(1))-4w`
`implies-w+4w=4(hc)/(lamda_(1))-(hc)/(lamda_(2))`
`implies 3w=hc((4)/(lamda_(1))-(1)/(lamda_(2))`
`w=(hc)/(3)((4)/(400xx10^(-9))-(1)/(250xx10^(9)))`
`=(hc)/(3)(10^(7)-(100xx10^(7))/(250))`
`(hc)/(3)xx10^(7)(1-0.4)`
`=0.2xx10^(7)J`
`thereforew=hcxx10^(6)J`