Question

The kinetic energy of the photoelectrons increases by $0.52 \, eV$ when the wavelength of incident light is changed from $500 \, nm$ to another wavelength which is approximately

• A. 700 nm
• B. 400 nm
• C. 1250 nm
• D. 1000 nm

Answer 1

Answer: (B)

Here, change in kinetic energy, $\Delta K = 0.52 \, eV, \lambda = 500 \, nm, \lambda_2 = ?$
We know , $ K_1 = \frac{hc}{\lambda_1 } = \phi $
$ K_2 = \frac{hc}{\lambda_2} - \phi$
$ \therefore \, K_1 = K_2 = hc \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right)$
or, $ - \Delta K = hc \left( \frac{1}{\lambda_1 } - \frac{1}{\lambda_2} \right) $
or, $ - 0.52 eV = \left(1242 eV nm\right) \left(\frac{1}{500 nm } - \frac{1}{\lambda_{2}}\right)$
or, $ \frac{-0.52}{1242} = \frac{1}{500} - \frac{1}{\lambda_{2}} $
or,$ \frac{1}{\lambda_{2}} = \frac{1}{500} + \frac{0.52}{1242} $
or, $ \lambda_{2} = 413 nm = 400 nm $