When radiation of 400 nm wavelength Is incident on photocell ,emitter produces photoelectron with maximum kinetic energy of 1.68 eV work function of emitter will be………. [hc=1240 eVnm]

Options

A. 3.09 eV
B. 1.42 eV
C. 1.51 eV
D. 1.68 V

Answer

B

Explanation

From Einstein.s equation of photoelectric effect,
`h_(f)=hf_(0)+(1)/(2)mv_(max)^(2)`
`therefore (h_(c))/(lambda)=phi+(1)/(2)mv_(max)^(2)`
`(1240)/(400)=phi+1.68`
`therefore phi=3.1-1.68 therefore phi=1.42 eV`