The number of revolutions/sec made by an electron in lind orbit is 8 times of the number of revolutions/sec made by an electron in `n^"th"` orbit. Give the value of "n".

Answer

To solve the problem, we need to find the value of "n" based on the given information about the revolutions per second made by an electron in different orbits.

### Step-by-Step Solution:

1. **Understanding Frequency**:
The frequency (f) of an electron in an orbit is defined as the number of revolutions per second. It can be expressed as:
\[ f = \frac{V}{C} \]
where \( V \) is the velocity of the electron and \( C \) is the circumference of the orbit.

2. **Circumference of the Orbit**:
The circumference \( C \) of an orbit is given by:
\[ C = 2\pi r \]
where \( r \) is the radius of the orbit.

3. **Velocity of the Electron**:
The velocity of the electron in the nth orbit can be derived from Bohr's model of the hydrogen atom:
\[ V_n = \frac{e^2}{h} \cdot \frac{1}{n} \]
where \( e \) is the charge of the electron and \( h \) is Planck's constant.

4. **Frequency in nth Orbit**:
Thus, the frequency of the electron in the nth orbit can be expressed as:
\[ f_n = \frac{V_n}{C_n} = \frac{V_n}{2\pi r_n} \]

5. **Comparing Frequencies**:
According to the problem, the frequency of the electron in the second orbit (n=2) is 8 times that in the nth orbit:
\[ f_2 = 8 f_n \]

6. **Using the Formula for Frequency**:
The ratio of frequencies can be expressed as:
\[ \frac{f_2}{f_n} = \frac{n^3}{2^3} \]
Substituting the known values:
\[ 8 = \frac{n^3}{2^3} \]

7. **Solving for n**:
Simplifying the equation:
\[ 8 = \frac{n^3}{8} \]
Multiplying both sides by 8:
\[ 64 = n^3 \]
Taking the cube root:
\[ n = 4 \]

### Final Answer:
The value of \( n \) is 4.