Question

The ionization energy of hydrogen is $13.6eV$. The energy of the photon released when an electron jumps from the first excited state $(n=2)$ to the ground state of a hydrogen atom is:

• A. $3.4eV$
• B. $4.53eV$
• C. $10.2eV$
• D. $13.6eV$

Answer 1

Answer: (C)

$10.2eV$

Transition of electron takes place from first excited state to ground state of Hydrogen atom.

$\therefore$ $n_1=1$ and $n_2=2$

Energy of photon released, $E=13.6(\frac{1}{n_1^2}-\frac{1}{n_2^2})$ $eV$

$E=13.6(\frac{1}{1^2}-\frac{1}{2^2})$$eV=13.6\times \frac{3}{4}eV=10.2$ $eV$