An electron jumps from the fourth orbit to the second orbit hydrogen atom. Given the Rydberg's constant `R = 10^(7) cm^(-1)`. The frequecny , in `Hz` , of the emitted radiation will be

Options

A. `(3)/(16) xx 10^(5)`
B. `(3)/(6) xx 10^(15)`
C. `(9)/(16) xx 10^(5)`
D. `(9)/(16) xx 10^(17)`

Explanation

To find the frequency of the emitted radiation when an electron jumps from the fourth orbit to the second orbit in a hydrogen atom, we can follow these steps:

### Step 1: Identify the transition levels
The electron transitions from the fourth orbit (n2 = 4) to the second orbit (n1 = 2).

### Step 2: Use the Rydberg formula
The Rydberg formula for the wavelength (λ) of the emitted radiation is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant.

### Step 3: Substitute the values into the formula
Given \( R = 10^7 \, \text{cm}^{-1} \), \( n_1 = 2 \), and \( n_2 = 4 \):
\[ \frac{1}{\lambda} = 10^7 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]

### Step 4: Calculate the fractions
Calculate \( \frac{1}{2^2} \) and \( \frac{1}{4^2} \):
\[ \frac{1}{2^2} = \frac{1}{4}, \quad \frac{1}{4^2} = \frac{1}{16} \]
Now, substituting these values:
\[ \frac{1}{\lambda} = 10^7 \left( \frac{1}{4} - \frac{1}{16} \right) \]

### Step 5: Find a common denominator and simplify
The common denominator for 4 and 16 is 16:
\[ \frac{1}{4} = \frac{4}{16} \]
Thus,
\[ \frac{1}{\lambda} = 10^7 \left( \frac{4}{16} - \frac{1}{16} \right) = 10^7 \left( \frac{3}{16} \right) \]
This simplifies to:
\[ \frac{1}{\lambda} = \frac{3 \times 10^7}{16} \]

### Step 6: Calculate λ
Now, taking the reciprocal to find λ:
\[ \lambda = \frac{16}{3 \times 10^7} \, \text{cm} \]

### Step 7: Calculate the frequency (ν)
The frequency (ν) can be calculated using the speed of light (c):
\[ \nu = \frac{c}{\lambda} \]
Given \( c = 3 \times 10^{10} \, \text{cm/s} \):
\[ \nu = \frac{3 \times 10^{10}}{\frac{16}{3 \times 10^7}} = 3 \times 10^{10} \times \frac{3 \times 10^7}{16} \]

### Step 8: Simplify the frequency calculation
\[ \nu = \frac{9 \times 10^{17}}{16} \, \text{Hz} \]

### Final Answer
Thus, the frequency of the emitted radiation is:
\[ \nu = \frac{9}{16} \times 10^{17} \, \text{Hz} \]