If in Hydrogen atom, an electron jumps from `n_(2) = 2` to `n_(1) = 1` in Bohr's orbit, then the value of wave number of the emitted photon will be `(R = 109700 cm^(-1))`

Options

A. `54850 cm^(-1)`
B. `82275 cm^(-1)`
C. `62875 cm^(-1)`
D. `10970 cm^(-1)`

Answer

To find the wave number of the emitted photon when an electron in a hydrogen atom jumps from \( n_2 = 2 \) to \( n_1 = 1 \), we can use the Rydberg formula for hydrogen:

\[ \bar{\nu} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( \bar{\nu} \) is the wave number,
- \( R \) is the Rydberg constant (given as \( 109700 \, \text{cm}^{-1} \)),
- \( n_1 \) is the lower energy level,
- \( n_2 \) is the higher energy level.

### Step 1: Identify the values of \( n_1 \) and \( n_2 \)
From the problem, we have:
- \( n_1 = 1 \)
- \( n_2 = 2 \)

### Step 2: Substitute the values into the Rydberg formula
Substituting the values into the formula gives:

\[ \bar{\nu} = 109700 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]

### Step 3: Calculate the fractions
Calculating the fractions:

\[ \frac{1}{1^2} = 1 \]
\[ \frac{1}{2^2} = \frac{1}{4} \]

### Step 4: Substitute the fractions back into the equation
Now substituting these values back into the equation:

\[ \bar{\nu} = 109700 \left( 1 - \frac{1}{4} \right) \]

### Step 5: Simplify the expression
Now simplify the expression inside the parentheses:

\[ 1 - \frac{1}{4} = \frac{4}{4} - \frac{1}{4} = \frac{3}{4} \]

### Step 6: Final calculation
Now we can calculate:

\[ \bar{\nu} = 109700 \times \frac{3}{4} \]

Calculating this gives:

\[ \bar{\nu} = 109700 \times 0.75 = 82275 \, \text{cm}^{-1} \]

### Conclusion
Thus, the wave number of the emitted photon when the electron jumps from \( n_2 = 2 \) to \( n_1 = 1 \) is:

\[ \bar{\nu} = 82275 \, \text{cm}^{-1} \]

### Answer
The correct answer is option 2: \( 82275 \, \text{cm}^{-1} \).

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