Question

If the energy of an electron in the first Bohr orbit of the H-atom is \(-2.18\times 10^{-18}\) J then the energy of the electron in the second orbit will be

• A. \(-2.18\times 10^{-18}\) J
• B. \(-4.36\times 10^{-18}\) J
• C. \(-0.545\times 10^{-18}\) J
• D. \(-0.273\times 10^{-18}\) J

Hint:

Use $E_n=\frac{E_1}{n^2}$.

Answer 1

The Correct Option is C

Step 1: Ratio
$\frac{E_2}{E_1}=\frac{1}{4}$, so $E_2=\frac14\times(-2.18\times10^{-18})=-5.45\times10^{-19}$ J.

Final Answer:
Option (C). \[ \boxed{\text{(C)}} \]