To which orbit the electron in H atom will jump on absorbing 12.1 eV energy ?

Answer

To determine to which orbit the electron in a hydrogen atom will jump upon absorbing 12.1 eV of energy, we can follow these steps:

### Step 1: Convert Energy from eV to Joules
The energy given is 12.1 eV. To convert this energy into Joules, we use the conversion factor where 1 eV = \(1.6 \times 10^{-19}\) Joules.

\[ \text{Energy in Joules} = 12.1 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.936 \times 10^{-18} \, \text{J} \]

### Step 2: Use the Rydberg Formula
The change in energy when an electron jumps between orbits in a hydrogen atom is given by the Rydberg formula:

\[ \Delta E = R_H \left( \frac{1}{n_i^2} - \frac{1}{n_f^2} \right) \]

Where:
- \(R_H\) is the Rydberg constant, approximately \(2.18 \times 10^{-18} \, \text{J}\).
- \(n_i\) is the initial orbit (for hydrogen, we can assume it starts from the ground state, so \(n_i = 1\)).
- \(n_f\) is the final orbit we need to find.

### Step 3: Substitute Known Values into the Formula
We know:
- \(\Delta E = 1.936 \times 10^{-18} \, \text{J}\)
- \(R_H = 2.18 \times 10^{-18} \, \text{J}\)
- \(n_i = 1\)

Substituting these values into the Rydberg formula:

\[ 1.936 \times 10^{-18} = 2.18 \times 10^{-18} \left( \frac{1}{1^2} - \frac{1}{n_f^2} \right) \]

### Step 4: Rearranging the Equation
We can rearrange the equation to isolate \(n_f\):

\[ \frac{1.936}{2.18} = 1 - \frac{1}{n_f^2} \]

Calculating the left side:

\[ 0.888 = 1 - \frac{1}{n_f^2} \]

### Step 5: Solve for \(n_f^2\)
Now, we can solve for \(\frac{1}{n_f^2}\):

\[ \frac{1}{n_f^2} = 1 - 0.888 = 0.112 \]

Taking the reciprocal gives:

\[ n_f^2 = \frac{1}{0.112} \approx 8.94 \]

### Step 6: Calculate \(n_f\)
Now, we take the square root to find \(n_f\):

\[ n_f \approx \sqrt{8.94} \approx 3 \]

### Conclusion
The electron in the hydrogen atom will jump to the **3rd orbit** upon absorbing 12.1 eV of energy.

---