In H-atom electron jumps from `3^(rd)` to `2^(nd)` energy level, the energy released is -

Options

A. `3.03 xx 10^-19 J//"atom"`
B. `1.03 xx 10^-19 J//"atom"`
C. `3.03 xx 10^-12 J//"atom"`
D. `6.06 xx 10^-19 J//"atom"`

Explanation

To find the energy released when an electron in a hydrogen atom jumps from the 3rd energy level (n=3) to the 2nd energy level (n=2), we can follow these steps:

### Step 1: Understand the Energy Levels
The energy of an electron in a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

### Step 2: Calculate the Energy at n=3
For \( n = 3 \):
\[ E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -\frac{13.6 \, \text{eV}}{9} = -1.51 \, \text{eV} \]

### Step 3: Calculate the Energy at n=2
For \( n = 2 \):
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]

### Step 4: Calculate the Energy Released
The energy released when the electron jumps from n=3 to n=2 is given by:
\[ \Delta E = E_2 - E_3 \]
Substituting the values we calculated:
\[ \Delta E = (-3.4 \, \text{eV}) - (-1.51 \, \text{eV}) \]
\[ \Delta E = -3.4 + 1.51 = -1.89 \, \text{eV} \]

### Step 5: Convert Energy to Joules
To convert the energy from electron volts to joules, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ \Delta E = -1.89 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} \]
\[ \Delta E \approx -3.02 \times 10^{-19} \, \text{J} \]

### Final Answer
The energy released when the electron jumps from the 3rd to the 2nd energy level in a hydrogen atom is approximately:
\[ \Delta E \approx -1.89 \, \text{eV} \text{ or } -3.02 \times 10^{-19} \, \text{J} \]

---