The energy required for transition in a Hydrogen atom from `1^(st)` energy level to the `2^(nd)` energy level is E. Then , which of the following transitions is possible for the same energy in Helium atom ?

Options

A. `2 rarr3`
B. `2rarr4`
C. `1rarr2`
D. `1rarr3`

Answer

To solve the problem of finding which transition in a Helium atom corresponds to the same energy \( E \) required for a transition from the first energy level to the second energy level in a Hydrogen atom, we can follow these steps:

### Step 1: Determine the Energy Transition in Hydrogen
The energy levels of a hydrogen atom are given by the formula:
\[ E_n = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \]
For Hydrogen, \( Z = 1 \).

The energy required for the transition from the first energy level (\( n_1 = 1 \)) to the second energy level (\( n_2 = 2 \)) is:
\[ E = E_2 - E_1 = \left(-\frac{1^2 \cdot 13.6}{2^2}\right) - \left(-\frac{1^2 \cdot 13.6}{1^2}\right) \]
Calculating this gives:
\[ E = \left(-\frac{13.6}{4}\right) - \left(-13.6\right) = -3.4 + 13.6 = 10.2 \, \text{eV} \]

### Step 2: Determine the Energy Transition in Helium
For Helium, \( Z = 2 \). The energy levels are given by:
\[ E_n = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} = -\frac{4 \cdot 13.6}{n^2} \]
We want to find \( n_1 \) and \( n_2 \) such that the energy difference corresponds to \( E = 10.2 \, \text{eV} \).

The energy required for a transition from \( n_1 \) to \( n_2 \) in Helium is:
\[ E' = E_{n_2} - E_{n_1} = \left(-\frac{4 \cdot 13.6}{n_2^2}\right) - \left(-\frac{4 \cdot 13.6}{n_1^2}\right) \]
This simplifies to:
\[ E' = 4 \cdot 13.6 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \]
Setting \( E' = 10.2 \, \text{eV} \):
\[ 4 \cdot 13.6 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) = 10.2 \]

### Step 3: Solve for \( n_1 \) and \( n_2 \)
Rearranging gives:
\[ \frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{10.2}{4 \cdot 13.6} \]
Calculating the right side:
\[ \frac{10.2}{54.4} = \frac{1}{5.333} \approx 0.1875 \]

Assuming \( n_1 = 2 \) and \( n_2 = 4 \):
\[ \frac{1}{2^2} - \frac{1}{4^2} = \frac{1}{4} - \frac{1}{16} = \frac{4}{16} - \frac{1}{16} = \frac{3}{16} \approx 0.1875 \]

Thus, the transition from \( n_1 = 2 \) to \( n_2 = 4 \) in Helium corresponds to the same energy \( E \) as the transition from \( n_1 = 1 \) to \( n_2 = 2 \) in Hydrogen.

### Conclusion
The possible transition in the Helium atom that corresponds to the same energy \( E \) is from the second energy level to the fourth energy level.