The frequency of radiations emitted when electron falls from `n = 4` to `n = 1` in `H-"atom"` would be (Given `E_1` for `H = 2.18 xx 10^-18 J "atom"^-1` and `h = 6.625 xx 10^-34 Js`.)

Options

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

Explanation

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

### Step 1: Calculate the energy difference between the two levels
The energy of an electron in a hydrogen atom at a specific energy level \( n \) is given by the formula:

\[ E_n = -\frac{E_1}{n^2} \]

where \( E_1 \) is the energy of the electron in the ground state (for hydrogen, \( E_1 = 2.18 \times 10^{-18} \, \text{J} \)).

#### Calculation:
1. Calculate \( E_4 \):
\[ E_4 = -\frac{E_1}{4^2} = -\frac{2.18 \times 10^{-18}}{16} = -1.3625 \times 10^{-19} \, \text{J} \]

2. Calculate \( E_1 \):
\[ E_1 = -2.18 \times 10^{-18} \, \text{J} \]

3. Find the energy difference \( \Delta E \):
\[ \Delta E = E_1 - E_4 = -2.18 \times 10^{-18} - (-1.3625 \times 10^{-19}) = -2.04375 \times 10^{-18} \, \text{J} \]

### Step 2: Use the energy difference to find the frequency
The frequency \( \nu \) of the emitted radiation can be calculated using Planck's equation:

\[ E = h \nu \]

where \( h \) is Planck's constant (\( h = 6.625 \times 10^{-34} \, \text{Js} \)).

#### Calculation:
Rearranging the equation gives:

\[ \nu = \frac{E}{h} \]

Substituting the values:

\[ \nu = \frac{2.04375 \times 10^{-18}}{6.625 \times 10^{-34}} \approx 3.08 \times 10^{15} \, \text{s}^{-1} \]

### Final Answer:
The frequency of the radiation emitted when the electron falls from \( n = 4 \) to \( n = 1 \) in a hydrogen atom is approximately:

\[ \nu \approx 3.08 \times 10^{15} \, \text{s}^{-1} \]