Question

In a photoelectric effect experiment the threshold wavelength of the light is $380nm$. If the wave length of incident light is $260nm$, the maximum kinetic energy of emitted electrons will be:
Given E (in eV) = $\frac{1237}{\lambda (innm)}$

• A. 1.5 eV
• B. 4.5 eV
• C. 15.1 eV
• D. 3.0 eV

Answer 1

Answer: (A)

$K_{max}=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}$
$\Rightarrow K_{max}=hc\left(\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right)$
$\Rightarrow K_{max}=\left(1237\right)\left(\frac{380-260}{380\times 260}\right)$
$=1.5eV$