Question

Light of frequency $8\times {10}^{15}Hz$ is incident on a substance of photoelectric work function 6.125 eV. The maximum kinetic energy of the emitted photoelectrons is

• A. 17 eV
• B. 22 eV
• C. 27 eV
• D. 37 eV

Answer 1

The correct option is C

27 eV

$E=hv=6.6\times {10}^{-34}\times 8\times {10}^{15}=5.28\times {10}^{-18}J=33eVByusingE=W_0+K_{max}\Rightarrow K_{max}=E-W_0$
$=33-6.125=27eV$