A hydrogen atom of mass m emits a photon corresponding to the fourth line of Brackett series and recoils. If R is the Rydberg's constant and h is the Planck's constant, then recoiling velocity is

Options

A. `(3Rh)/(36m)`
B. `(3Rh)/(64m)`
C. `(7Rh)/(120m)`
D. `(7Rh)/(3m)`

Answer

To find the recoiling velocity of a hydrogen atom that emits a photon corresponding to the fourth line of the Brackett series, we can follow these steps:

### Step 1: Identify the transition for the fourth line of the Brackett series
The Brackett series corresponds to transitions where the final energy level (n1) is 4. The fourth line means the transition from n2 = 8 to n1 = 4.

### Step 2: Use the Rydberg formula to find the wavelength of the emitted photon
The Rydberg formula is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Substituting \( n_1 = 4 \) and \( n_2 = 8 \):
\[ \frac{1}{\lambda} = R \left( \frac{1}{4^2} - \frac{1}{8^2} \right) = R \left( \frac{1}{16} - \frac{1}{64} \right) \]

### Step 3: Simplify the expression for \( \frac{1}{\lambda} \)
Calculating the fractions:
\[ \frac{1}{16} - \frac{1}{64} = \frac{4}{64} - \frac{1}{64} = \frac{3}{64} \]
Thus,
\[ \frac{1}{\lambda} = R \cdot \frac{3}{64} \]
This gives:
\[ \lambda = \frac{64}{3R} \]

### Step 4: Calculate the momentum of the emitted photon
The momentum \( p \) of a photon is given by:
\[ p = \frac{h}{\lambda} \]
Substituting for \( \lambda \):
\[ p = \frac{h}{\frac{64}{3R}} = \frac{3hR}{64} \]

### Step 5: Apply conservation of momentum
According to the conservation of momentum, the momentum of the recoiling hydrogen atom must equal the momentum of the emitted photon:
\[ p_H = p \]
Let \( v \) be the recoiling velocity of the hydrogen atom:
\[ mv = \frac{3hR}{64} \]

### Step 6: Solve for the recoiling velocity \( v \)
Rearranging the equation gives:
\[ v = \frac{3hR}{64m} \]

### Final Answer
The recoiling velocity of the hydrogen atom is:
\[ v = \frac{3hR}{64m} \]

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