Question

A photon of energy $8eV$ is incident on metal surface of threshold frequency $1.6\times {10}^{15}Hz$. The maximum kinetic energy of the photoelectrons emitted (in eV) (Take $h=6\times {10}^{-34}Js)$.

• A. $1.4eV$
• B. $2.4eV$
• C. $4.8eV$
• D. $0.8eV$
• E. $7eV$

Answer 1

The correct option is A $1.4eV$

As we know,

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