Two electrons are revolving around a nucleus at distances 'r' and '4r' ratio of their periods is

Options

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

Answer

To solve the problem of finding the ratio of the periods of two electrons revolving around a nucleus at distances 'r' and '4r', we can follow these steps:

### Step 1: Understand the relationship between radius and period
The time period \( T \) of an electron revolving in a circular orbit is given by:
\[ T = \frac{2\pi r}{v} \]
where \( r \) is the radius of the orbit and \( v \) is the velocity of the electron.

### Step 2: Establish the relationship between velocity and radius
For an electron in a circular orbit, the centripetal force is provided by the electrostatic force between the electron and the nucleus. According to Bohr's model, the velocity \( v \) of the electron is inversely proportional to the principal quantum number \( n \):
\[ v \propto \frac{1}{n} \]
Also, the radius \( r \) is directly proportional to \( n^2 \):
\[ r \propto n^2 \]

### Step 3: Relate the time period to quantum number
Substituting the relationships into the equation for the time period:
\[ T \propto \frac{r}{v} \propto \frac{n^2}{\frac{1}{n}} = n^3 \]
Thus, we can conclude:
\[ T \propto n^3 \]

### Step 4: Establish the relationship between periods and radii
Now, we can relate the periods of the two electrons. Let \( T_1 \) be the period of the electron at radius \( r \) and \( T_2 \) be the period of the electron at radius \( 4r \). Let \( n_1 \) and \( n_2 \) be the quantum numbers corresponding to these radii:
- For \( r \): \( n_1^2 \propto r \)
- For \( 4r \): \( n_2^2 \propto 4r \)

Since \( n_2^2 = 4n_1^2 \), we have:
\[ n_2 = 2n_1 \]

### Step 5: Calculate the ratio of the periods
Using the relationship \( T \propto n^3 \):
\[ \frac{T_1}{T_2} = \frac{n_1^3}{n_2^3} = \frac{n_1^3}{(2n_1)^3} = \frac{n_1^3}{8n_1^3} = \frac{1}{8} \]

### Final Answer
Thus, the ratio of their periods is:
\[ T_1 : T_2 = 1 : 8 \]