Question

In a collection of H-atoms, all the electrons jump from n = 5 to ground level finally (directly or indirectly), without emitting any line in Balmer series. The number of possible different radiations is :

• A. 10
• B. 8
• C. 7
• D. 6

Answer 1

Answer: (C)

7

Answer 2

The total number of possible spectral lines emitted when electrons transition from a higher energy level $n_i$ to any lower energy level $n_f$ is given by the formula $\binom{n_i}{2}$. In this problem, electrons start from $n=5$ and can cascade down to $n=1$, involving levels {1, 2, 3, 4, 5}. The total number of possible distinct radiations is $\binom{5}{2} = \frac{5 \times 4}{2} = 10$.

The Balmer series corresponds to transitions where the final energy level is $n_f = 2$. These transitions are:

1. $n_i = 5 \to n_f = 2$
2. $n_i = 4 \to n_f = 2$
3. $n_i = 3 \to n_f = 2$

There are 3 such transitions that produce Balmer lines. The problem states that no line in the Balmer series is emitted, which means these 3 transitions are forbidden.

Therefore, the number of possible different radiations is the total number of possible transitions minus the number of forbidden (Balmer) transitions: Number of allowed radiations = (Total possible radiations) - (Number of Balmer radiations) Number of allowed radiations = $10 - 3 = 7$.