Question

A photon of energy 8 eV is incident on a metal surface of threshold frequency $1.6\times {10}^{15}Hz$, then the maximum kinetic energy of photoelectrons emitted is $(h=6.6\times {10}^{-34}Js)$

• A. 4.8 eV
• B. 2.4 eV
• C. 1.4 eV
• D. 0.8 eV

Answer 1

The correct option is C 1.4 eV

Work function $W_0=h{v}_0=6.6\times {10}^{-34}\times 1.6\times {10}^{15}$
$=1.056\times {10}^{-18}J=6.6eV$
$FromE=W_0+K_{max}\Rightarrow K_{max}=E-W_0=1.4eV$