Question

Find the energy of third stationary orbit of Bohr's model of hydrogen atom if the energy of ground state is \(-E\) J

• A. \(-\frac{E}{2}\) J
• B. \(-\frac{E}{4}\) J
• C. \(-\frac{E}{9}\) J
• D. \(-\frac{E}{16}\) J

Hint:

In a hydrogen atom the orbit energy varies as 1/n squared.

Answer 1

The Correct Option is C

Step 1: Set up the ratio:
Since $E_n \propto \frac{1}{n^2}$, we can write $\frac{E_3}{E_1} = \frac{1^2}{3^2}$.

Step 2: Substitute:
\[ E_3 = E_1\times\frac{1}{9} = (-E)\times\frac{1}{9} = -\frac{E}{9} \]
The third orbit is less tightly bound than the ground state, so its energy is closer to zero. This matches option (C).

Final Answer:
Energy of the third orbit is $-\frac{E}{9}$ J. \[ \boxed{-\frac{E}{9}\text{ J}} \]