Question

The ratio of energies of photons produced due to transition of electron of hydrogen atom from its (i) third to 2nd energy level and (ii) highest energy level to 3rd level is ______.

• A. 3 : 2
• B. 5 : 4
• C. 5 : 3
• D. 8 : 3

Hint:

Transitions ending at $n=2$ are the Balmer series (visible light). Transitions ending at $n=3$ are the Paschen series (infrared). Higher energy drops generally yield larger numerators!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We must use the Rydberg energy formula for the hydrogen atom to calculate the energies of two specific photon emissions and find their ratio.

Step 2: Detailed Explanation:
The energy of a photon emitted during an electron transition from a higher level $n_2$ to a lower level $n_1$ in a hydrogen atom is given by:
$E = 13.6 \text{ eV} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
Case (i): Transition from 3rd to 2nd energy level
Here, $n_2 = 3$ and $n_1 = 2$.
$E_1 = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right)$
$E_1 = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right)$
$E_1 = 13.6 \left( \frac{9 - 4}{36} \right)$
$E_1 = 13.6 \left( \frac{5}{36} \right)$
Case (ii): Transition from highest level to 3rd level
The "highest energy level" mathematically means $n_2 = \infty$. The destination is $n_1 = 3$.
$E_2 = 13.6 \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right)$
Since $\frac{1}{\infty} = 0$:
$E_2 = 13.6 \left( \frac{1}{9} - 0 \right)$
$E_2 = 13.6 \left( \frac{1}{9} \right)$
To make the ratio easier to compute, express this with a denominator of 36:
$E_2 = 13.6 \left( \frac{4}{36} \right)$
Find the ratio $E_1 / E_2$:
$\text{Ratio} = \frac{13.6 \times \frac{5}{36}}{13.6 \times \frac{4}{36}}$
The constants $13.6$ and the denominators $36$ perfectly cancel out:
$\text{Ratio} = \frac{5}{4}$

Step 3: Final Answer:
The ratio is 5 : 4, matching option (b).