In Bohr's model of the hydrogen atom the ratio between the period of revolution of an electron in the orbit of `n=1` to the period of the revolution of the electron in the orbit `n=2` is :-
(a). `1 : 2`
(b). `2 : 1`
(c). `1 : 4`
(d). `1 : 8`

Options

A. `2:1`
B. `1:2`
C. `1:4`
D. `1:8`

Answer

To solve the problem, we will use Bohr's model of the hydrogen atom, which states that the time period of an electron in a particular orbit is directly proportional to the cube of the principal quantum number (n) of that orbit.

### Step-by-Step Solution:

1. **Understanding the Relationship**: According to Bohr's model, the time period \( T \) of an electron in orbit is given by the relation:
\[ T \propto n^3 \]
This means that the time period increases with the cube of the principal quantum number.

2. **Setting Up the Ratio**: We need to find the ratio of the time periods for the orbits with \( n = 1 \) and \( n = 2 \). We denote:
- \( T_1 \) as the time period for \( n = 1 \)
- \( T_2 \) as the time period for \( n = 2 \)

Using the proportionality, we can write:
\[ \frac{T_1}{T_2} = \frac{n_1^3}{n_2^3} \]

3. **Substituting the Values**: Now, we substitute \( n_1 = 1 \) and \( n_2 = 2 \):
\[ \frac{T_1}{T_2} = \frac{1^3}{2^3} = \frac{1}{8} \]

4. **Final Ratio**: This means that:
\[ T_1 : T_2 = 1 : 8 \]

5. **Conclusion**: Therefore, the ratio between the period of revolution of an electron in the orbit of \( n = 1 \) to the period of the revolution of the electron in the orbit \( n = 2 \) is \( 1 : 8 \).

### Answer:
The correct option is (d) \( 1 : 8 \).