JEE Physics 2024 Question with Solution

Given: An electron in a hydrogen atom has energy $E_n = -0.85 \, \text{eV}$ in an excited state.

Find: The maximum number of allowed transitions to lower energy levels.

Use the Bohr model relation for hydrogen:

$$E_n = -\frac{13.6}{n^2} \, \text{eV}$$

Substitute the given energy:

$$-0.85 = -\frac{13.6}{n^2}$$

So,

$$n^2 = \frac{13.6}{0.85} = 16$$

Hence,

$$n = 4$$

Now use the formula for the maximum number of allowed transitions from level $n$:

$$\text{Number of transitions} = \frac{n(n-1)}{2}$$

Substituting $n = 4$,

$$\text{Number of transitions} = \frac{4(4-1)}{2} = \frac{12}{2} = 6$$

Therefore, the maximum number of allowed transitions is $6$. The correct option is C.

Given: $E_n = -0.85 \, \text{eV}$.

Find: The maximum number of transitions.

First identify the orbit using

$$E_n = -\frac{13.6}{n^2} \, \text{eV}$$

Since

$$\frac{13.6}{0.85} = 16$$

we get

$$n = 4$$

For an electron in the $4$th level, the maximum number of distinct downward transitions is directly

$$\frac{n(n-1)}{2} = \frac{4 \times 3}{2} = 6$$

This shortcut works because every pair of energy levels among the lower reachable levels gives one possible emission line. Therefore, the correct option is C.