The energy of ground state in hydrogen atom is \( -13.6\ \mathrm{eV} \). If the electron jumps to ground state from the \( 4^{\mathrm{th}} \) excited state, then the wavelength of the photon emitted will be approximately
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 952 \) Å
In the Bohr model of hydrogen, \( E_{n} = -13.6/n^{2}\ \mathrm{eV} \). The 4th excited state is \( n = 5 \) (the ground state is \( n = 1 \), the first excited state \( n = 2 \)).
\( \displaystyle \Delta E = 13.6\left(1 - \frac{1}{25}\right) = 13.6 \times \frac{24}{25} = 13.056\ \mathrm{eV} \)
\( \displaystyle \lambda = \frac{hc}{\Delta E} = \frac{(6.63\times10^{-34})(3\times10^{8})}{13.056 \times 1.6\times10^{-19}}\ \mathrm{m} \approx 9.52\times10^{-8}\ \mathrm{m} \)
\( \lambda \approx 952 \) Å