Question

The ionization potential of a hydrogen atom is 13.6 eV. The wavelength of the energy radiation required for the ionization of H-atom is:

• A. 1911 nm
• B. 912 nm
• C. 68 nm
• D. 91.2 nm

Answer 1

The correct option is D 91.2 nm

$E=\frac{hc}{\lambda }$
$13.6eV=\frac{hc}{\lambda }$
$\lambda =\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{13.6\times 1.6\times {10}^{-19}J}=91.2nm$.