The frequency of radiation emiited when the electron falls n =4 to n=1 in a hydrogen atom will be ( given ionization energy of `H= 2.18 xx 10 ^(-18)J "atom "^(-1) and h= 6.625 xx 10 ^(-34)Js)`

Options

A. `1.54xx 10^(15)s^(-1)`
B. `1.03 xx 10^(15)s^(-1)`
C. `3.08xx 10^(15 )s^(-1)`
D. `2.00xx10^(15) s^(-1)`

Answer

To calculate the frequency of radiation emitted when an electron falls from the n=4 state to the n=1 state in a hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Ionization Energy**:
The ionization energy (E1) for hydrogen is given as \( E_1 = -2.18 \times 10^{-18} \, \text{J} \).

2. **Calculate Energy at n=4**:
The energy of the electron in the n=4 state (\( E_4 \)) can be calculated using the formula:
\[ E_n = \frac{E_1}{n^2} \]
Substituting \( n = 4 \):
\[ E_4 = \frac{-2.18 \times 10^{-18}}{4^2} = \frac{-2.18 \times 10^{-18}}{16} = -1.36 \times 10^{-19} \, \text{J} \]

3. **Calculate the Change in Energy (\( \Delta E \))**:
The change in energy when the electron transitions from n=4 to n=1 is given by:
\[ \Delta E = E_1 - E_4 \]
Substituting the values:
\[ \Delta E = -2.18 \times 10^{-18} - (-1.36 \times 10^{-19}) = -2.18 \times 10^{-18} + 1.36 \times 10^{-19} \]
\[ \Delta E = -2.04 \times 10^{-18} \, \text{J} \]

4. **Calculate the Frequency (\( \nu \))**:
The frequency of the emitted radiation can be calculated using the formula:
\[ \nu = \frac{\Delta E}{h} \]
where \( h = 6.625 \times 10^{-34} \, \text{J s} \). Substituting the values:
\[ \nu = \frac{2.04 \times 10^{-18}}{6.625 \times 10^{-34}} \approx 3.08 \times 10^{15} \, \text{s}^{-1} \]

5. **Final Answer**:
The frequency of radiation emitted when the electron falls from n=4 to n=1 in a hydrogen atom is approximately \( 3.08 \times 10^{15} \, \text{s}^{-1} \).