if electron falls from n=3 to n=2, then emitted energy is

Options

A. 10.2 eV
B. 12.09 eV
C. 1.9 eV
D. 0.65 eV

Explanation

To find the emitted energy when an electron falls from n=3 to n=2, we can use the formula for the energy of an electron in a hydrogen atom, which is given by:

\[ E_n = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]

where:
- \(E_n\) is the energy of the electron in the nth orbit,
- \(Z\) is the atomic number (for hydrogen, \(Z=1\)),
- \(n\) is the principal quantum number.

### Step 1: Calculate the energy at n=3
Using the formula, we can calculate the energy when \(n=3\):

\[ E_3 = -\frac{13.6 \, \text{eV} \cdot 1^2}{3^2} = -\frac{13.6 \, \text{eV}}{9} = -1.51 \, \text{eV} \]

### Step 2: Calculate the energy at n=2
Next, we calculate the energy when \(n=2\):

\[ E_2 = -\frac{13.6 \, \text{eV} \cdot 1^2}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]

### Step 3: Calculate the emitted energy
The emitted energy when the electron falls from n=3 to n=2 is the difference in energy between these two states:

\[ \text{Emitted Energy} = E_2 - E_3 \]

Substituting the values we calculated:

\[ \text{Emitted Energy} = (-3.4 \, \text{eV}) - (-1.51 \, \text{eV}) = -3.4 \, \text{eV} + 1.51 \, \text{eV} = -1.89 \, \text{eV} \]

Since energy emitted is a positive quantity, we take the absolute value:

\[ \text{Emitted Energy} = 1.89 \, \text{eV} \]

### Final Answer
The emitted energy when the electron falls from n=3 to n=2 is **1.89 eV**.

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