Consider a metal exposed to light of wavelength 600nm. The maximum energy of the electrons doubles when light of wavelength 400nm is used. Find the work function in eV.

Options

A. 1.1 ev
B. 1.03 eV
C. 1.5 eV
D. 1.8 eV

Answer

B

Explanation

`(K_("max"))_(1)=(hc)/(lambda_(1))-W_(0)`
and `(K_("max"))_(2)=(hc)/(lambda_(2))-W_(0)`
where `W_(0)` is the work function .
`:'(K_("max"))_(2)=2(K_("max"))_(1)` (given)
`:.((hc)/(lambda_(2))-W_(0))=2((hc)/(lambda_(1))-W_(0))`
But `hc=1240 `eV nm
`:.((1240)/(400)-W_(0))=2((1240)/(600)-W_(0))`
`=2((1240)/(600))-2W_(0)`
`:.W_(0)=(2xx1240)/(600)-(1240)/(400)=1240[(2)/(600)-(1)/(400)]`
`=(1240)/(1200)=1.03 eV`