When ultraviolet radiation of a certain frequency falls on a potassium target, the photoelectrons released can be stopped completely by a retarding potential of 0.6 V. If the frequency of the radiation is increased by 10%, this stopping potential rises to 0.9 V. The work function of potassium is

Options

A. 2.0 eV
B. 2.4 eV
C. 3.0 eV
D. 2.8 eV

Answer

B

Explanation

`KE_("max") =exxV_0`
`= exx0.6 = 0.6 eV`
Einstein photoelectric equation
`(KE)_("max") = hf -phi`
`implies E=hf =0.6phi" "...(i)`
`(KE_("max"))= exxV_0= exx0.9 = 0.9eV`
`1.1 E = 0.9 +phi`
By eq. `(ii)/(i) ,implies 1.1 = (0.9 +phi)/(0.6+phi)impliesphi=2.4eV`