An electrons of a stationary hydrogen aton passes form the fifth enegry level to the ground level. The velocity that the atom acquired as a result of photon emission will be
`(m` is the mass of the electron, `R`, Rydberg constanrt and `h`, Planck's constant)

Options

A. `(24m)/(25hR)`
B. `(24hR)/(25m)`
C. `(25hR)/(24m)`
D. `(25m)/(24hR)`

Explanation

To find the velocity acquired by a hydrogen atom when an electron transitions from the fifth energy level to the ground level, we can follow these steps:

### Step 1: Determine the Energy Difference
The energy difference (ΔE) between the two levels can be calculated using the Rydberg formula for hydrogen:

\[ \Delta E = hR \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( h \) is Planck's constant,
- \( R \) is the Rydberg constant,
- \( n_1 \) is the lower energy level (ground state, \( n_1 = 1 \)),
- \( n_2 \) is the higher energy level (fifth level, \( n_2 = 5 \)).

Substituting the values:

\[ \Delta E = hR \left( \frac{1}{1^2} - \frac{1}{5^2} \right) = hR \left( 1 - \frac{1}{25} \right) = hR \left( \frac{24}{25} \right) \]

### Step 2: Calculate the Momentum of the Photon Emitted
The momentum (p) of the emitted photon can be expressed as:

\[ p = \frac{\Delta E}{c} \]

Where \( c \) is the speed of light. Substituting the expression for ΔE:

\[ p = \frac{hR \left( \frac{24}{25} \right)}{c} \]

### Step 3: Relate Momentum to the Velocity of the Hydrogen Atom
The momentum of the hydrogen atom (which is initially at rest) after the photon is emitted can be expressed as:

\[ p = mv \]

Where:
- \( m \) is the mass of the hydrogen atom (considering the mass of the electron for simplicity),
- \( v \) is the velocity of the hydrogen atom.

Setting the two expressions for momentum equal gives:

\[ mv = \frac{hR \left( \frac{24}{25} \right)}{c} \]

### Step 4: Solve for the Velocity
Now, we can solve for the velocity \( v \):

\[ v = \frac{hR \left( \frac{24}{25} \right)}{mc} \]

### Final Expression
Thus, the velocity acquired by the hydrogen atom as a result of the photon emission when the electron transitions from the fifth energy level to the ground level is:

\[ v = \frac{24hR}{25m} \]