The recoil speed of hydrogen atom after it emits a photon in going from `n = 2 state to n = 1` state is nearly `[Take R_(oo) = 1.1 xx 10^(7) m and h = 6.63 xx 10^(-34) J s]`

Options

A. `1.5 m s^(-1)`
B. `3.3 m s^(-1)`
C. `4.5 m s^(-1)`
D. `6.6 m s^(-1)`

Explanation

To find the recoil speed of a hydrogen atom after it emits a photon while transitioning from the n=2 state to the n=1 state, we can follow these steps:

### Step 1: Calculate the energy difference (ΔE) between the two energy levels.

The energy of a photon emitted during a transition between two energy levels in a hydrogen atom can be calculated using the formula:

\[ \Delta E = E_2 - E_1 = h \cdot f = \frac{h \cdot c}{\lambda} \]

Where:
- \( h \) is Planck's constant (\(6.63 \times 10^{-34} \, \text{J s}\))
- \( c \) is the speed of light (\(3 \times 10^8 \, \text{m/s}\))
- \( \lambda \) is the wavelength of the emitted photon.

### Step 2: Calculate the wavelength (λ) using the Rydberg formula.

The wavelength can be calculated using the Rydberg formula:

\[ \frac{1}{\lambda} = R_{\infty} \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]

Where:
- \( R_{\infty} = 1.1 \times 10^7 \, \text{m}^{-1} \)
- \( n_f = 1 \) (final state)
- \( n_i = 2 \) (initial state)

Substituting the values:

\[ \frac{1}{\lambda} = 1.1 \times 10^7 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 1.1 \times 10^7 \left( 1 - \frac{1}{4} \right) = 1.1 \times 10^7 \left( \frac{3}{4} \right) \]

Calculating this gives:

\[ \frac{1}{\lambda} = 0.825 \times 10^7 \, \text{m}^{-1} \implies \lambda = \frac{1}{0.825 \times 10^7} \approx 1.21 \times 10^{-8} \, \text{m} \]

### Step 3: Calculate the energy difference (ΔE).

Now substituting \( \lambda \) back into the energy formula:

\[ \Delta E = \frac{h \cdot c}{\lambda} = \frac{(6.63 \times 10^{-34}) \cdot (3 \times 10^8)}{1.21 \times 10^{-8}} \]

Calculating this gives:

\[ \Delta E \approx 1.64 \times 10^{-19} \, \text{J} \]

### Step 4: Calculate the recoil speed (v) of the hydrogen atom.

Using the conservation of momentum, the recoil momentum of the hydrogen atom is equal to the momentum of the emitted photon:

\[ m \cdot v = \frac{\Delta E}{c} \]

Where:
- \( m \) is the mass of the hydrogen atom (\(1.67 \times 10^{-27} \, \text{kg}\))
- \( v \) is the recoil speed we want to find.

Rearranging gives:

\[ v = \frac{\Delta E}{m \cdot c} \]

Substituting the values:

\[ v = \frac{1.64 \times 10^{-19}}{(1.67 \times 10^{-27}) \cdot (3 \times 10^8)} \]

Calculating this gives:

\[ v \approx 3.3 \, \text{m/s} \]

### Final Answer:

The recoil speed of the hydrogen atom after emitting a photon is approximately **3.3 m/s**.

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