Question

In Hydrogen atom, an electron jumped from an orbit of radius 2592.1 pm to another orbit of radius 211.6 pm. What is the energy difference (in J) between these two states?

• A. $\approx2.18\times10^{-18}$
• B. $5\times10^{-19}$
• C. $5\times10^{-20}$
• D. $5\times10^{-18}$

Hint:

Radius $r_n \propto n^2$; energy $E_n \propto -1/n^2$. For transitions, $\Delta E>0$ for emission (higher to lower n). Convert eV to J using $1.6 \times 10^{-19}$. Approximate calculations for options.

Answer 1

The Correct Option is B

1. In the hydrogen atom, orbital radius $r_n = 52.9 n^2$ pm, where $n$ is the principal quantum number.
2. For larger radius 2592.1 pm: $n_2^2 = 2592.1 / 52.9 \approx 49$, so $n_2 = 7$.
3. For smaller radius 211.6 pm: $n_1^2 = 211.6 / 52.9 \approx 4$, so $n_1 = 2$.
4. Energy of level $E_n = -13.6 / n^2$ eV. Difference $\Delta E = 13.6 (1/n_1^2 - 1/n_2^2) = 13.6 (1/4 - 1/49) \approx 13.6 \times 0.2296 \approx 3.12$ eV.
5. Convert to joules: $3.12 \times 1.602 \times 10^{-19} \approx 5 \times 10^{-19}$ J.
6. Therefore, the correct option is (2) $5\times10^{-19}$.