Question

If the electron falls from $n=5$ to $n=4$ in the H-atom, then emitted energy is

• A. $0.306eV$
• B. $12.09eV$
• C. $1.89eV$
• D. $0.65eV$

Answer 1

Answer: (A)

$E_n=-\frac{13.6}{n^2}eV$

$E_5-E_4=13.6\left(\frac{1}{4^2}-\frac{1}{5^2}\right)eV$

$=13.6\left(\frac{1}{16}-\frac{1}{25}\right)=0.306eV$