Question

The energy (in J) released when an excited electron of 5$^{\text{th}}$ orbit of hydrogen atom returns to its ground state is

• A. \(2.091 \times 10^{-18}\)
• B. \(4.182 \times 10^{-18}\)
• C. \(6.273 \times 10^{-18}\)
• D. \(8.364 \times 10^{-18}\)

Hint:

For hydrogen atom transitions, use $\Delta E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ eV and convert eV to Joules by multiplying with $1.602 \times 10^{-19}$.

Answer 1

The Correct Option is A

Step 1: Write the formula for energy released during electron transition in hydrogen atom.
The energy released when an electron jumps from the $n^{\text{th}}$ orbit to the ground state ($n=1$) is: \[ \Delta E = E_1 - E_n = -13.6 \left(\frac{1}{1^2} - \frac{1}{n^2}\right) \text{ eV} \] For $n = 5$: \[ \Delta E = -13.6 \left(1 - \frac{1}{25}\right) = -13.6 \left(\frac{24}{25}\right) = -13.06 \text{ eV} \] The negative sign indicates energy is released.
Step 2: Convert energy from eV to Joules.
1 eV $= 1.602 \times 10^{-19}$ J \[ \Delta E = 13.06 \times 1.602 \times 10^{-19} \text{ J} \] \[ \Delta E \approx 20.91 \times 10^{-19} \text{ J} \] \[ \Delta E \approx 2.091 \times 10^{-18} \text{ J} \] Final Answer: \[ \boxed{2.091 \times 10^{-18} \text{ J}} \]