Question

Radiational wave length $\lambda$=124 nm falls on a metallic surface. Then the kinetic energy of the ejected photo electron(s) can be : (Given that threshold wavelength (${\lambda }_0$)=248 nm)

• A. 1 eV
• B. 2 eV
• C. 3 eV
• D. 5 eV

Answer 1

The correct option is D 5 eV

We know,

$KE=hv-h{v}_0$

$=hc[\frac{1}{\lambda }-\frac{1}{{\lambda }_0}]$

$=6.626\times {10}^{-34}\times 3\times {10}^8\times [\frac{1}{124\times {10}^{-9}}-\frac{1}{248\times {10}^{-9}}]$

$=0.08015\times {10}^{-17}$

$=8.015\times {10}^{-19}J$

$1.6\times {10}^{-19}J=1eV$

$8.015\times {10}^{-19}J=\frac{1}{1.6\times {10}^{-19}}\times 8.015\times {10}^{-19}$

$=4.74eV$

Thus, the answer is close to $5eV$.