A Bohr's hydrogen atom undergoes a transition `n = 5 to n = 4` and emits a photon of frequency `f`. Frequency of circular motion of electron in `n = 4 orbit is f_(4)`. The ratio `f//f_(4)` is found to be `18//5 m`. State the value of `m`.

Explanation

To solve the problem, we need to analyze the transition of a hydrogen atom from the energy level \( n = 5 \) to \( n = 4 \) and find the ratio of the emitted photon frequency \( f \) to the frequency of circular motion of the electron in the \( n = 4 \) orbit \( f_4 \).

### Step-by-Step Solution:

1. **Understanding Energy Levels**:
The energy levels of a hydrogen atom are given by the formula:
\[ E_n = -\frac{m z^2 e^4}{8 \epsilon_0^2 h^2 n^2} \]
where \( m \) is the mass of the electron, \( z \) is the atomic number (1 for hydrogen), \( e \) is the charge of the electron, \( \epsilon_0 \) is the permittivity of free space, and \( h \) is Planck's constant.

2. **Calculating Energy Difference**:
The energy difference between the levels \( n = 5 \) and \( n = 4 \) is given by:
\[ \Delta E = E_4 - E_5 = \left(-\frac{m z^2 e^4}{8 \epsilon_0^2 h^2 (4^2)}\right) - \left(-\frac{m z^2 e^4}{8 \epsilon_0^2 h^2 (5^2)}\right) \]
Simplifying this gives:
\[ \Delta E = \frac{m z^2 e^4}{8 \epsilon_0^2 h^2} \left(\frac{1}{25} - \frac{1}{16}\right) \]

3. **Finding the Frequency of the Photon**:
The frequency \( f \) of the emitted photon is related to the energy difference by:
\[ f = \frac{\Delta E}{h} \]
Substituting the expression for \( \Delta E \):
\[ f = \frac{m z^2 e^4}{8 \epsilon_0^2 h^3} \left(\frac{1}{25} - \frac{1}{16}\right) \]
The term \( \left(\frac{1}{25} - \frac{1}{16}\right) \) can be calculated as:
\[ \frac{1}{25} - \frac{1}{16} = \frac{16 - 25}{400} = -\frac{9}{400} \]
Therefore:
\[ f = \frac{m z^2 e^4}{8 \epsilon_0^2 h^3} \cdot \frac{9}{400} \]

4. **Calculating \( f_4 \)**:
The frequency of circular motion of the electron in the \( n = 4 \) orbit is given by:
\[ f_4 = \frac{m z^2 e^4}{8 \epsilon_0^2 h^3} \cdot \frac{1}{n^2} = \frac{m z^2 e^4}{8 \epsilon_0^2 h^3} \cdot \frac{1}{16} \]

5. **Finding the Ratio \( \frac{f}{f_4} \)**:
Now we can find the ratio:
\[ \frac{f}{f_4} = \frac{\frac{m z^2 e^4}{8 \epsilon_0^2 h^3} \cdot \frac{9}{400}}{\frac{m z^2 e^4}{8 \epsilon_0^2 h^3} \cdot \frac{1}{16}} = \frac{9}{400} \cdot 16 = \frac{144}{400} = \frac{36}{100} = \frac{18}{50} \]

6. **Setting the Ratio Equal to Given Value**:
We are given that:
\[ \frac{f}{f_4} = \frac{18}{5m} \]
Setting the two ratios equal gives:
\[ \frac{18}{50} = \frac{18}{5m} \]

7. **Solving for \( m \)**:
Cross-multiplying:
\[ 18 \cdot 5m = 18 \cdot 50 \]
Simplifying:
\[ 5m = 50 \implies m = 10 \]

### Final Answer:
The value of \( m \) is \( 10 \).