Question

Hydrogen atom makes a transition from the third excited state to the first excited state. The energy of the proton emitted in the process is:

• A. $1.89eV$
• B. $2.55eV$
• C. $12.09eV$
• D. $12.75eV$

Answer 1

The correct option is B $2.55eV$

Energy released $=13.6[\frac{1}{n{}_2^2}-\frac{1}{n{}_1^2}]$

$=13.6[\frac{1}{2^2}-\frac{1}{4^2}]$

Energy released $=13.6\times (\frac{1}{2^2}-\frac{1}{4^2})=2.55eV$