Time period of revolution of an electron in `n^(th)` orbit in a hydrogen like atom is given by ` T = (T_(0)n_(a))/ (Z^(b)) , Z = `atomic number

Options

A. `T_(0) = 1.5 xx 10^(-16)s, a = 3`
B. `T_(0) = 6.6 xx 10^(15)s, a = 3`
C. `T_(0) = 1.51 xx 10^(-16)s, b = 3`
D. `T_(0) = 6.6 xx 10^(15)s, b = 3`

Answer

To find the values of \( T_0 \), \( A \), and \( b \) in the equation for the time period of revolution of an electron in the \( n^{th} \) orbit of a hydrogen-like atom, we can follow these steps:

### Step 1: Write the formula for time period
The time period \( T \) of an electron in the \( n^{th} \) orbit is given by:
\[ T = \frac{T_0 n^A}{Z^b} \]
where \( Z \) is the atomic number.

### Step 2: Use Bohr's model to find expressions for radius and velocity
From Bohr's model:
- The radius \( R \) of the \( n^{th} \) orbit is given by:
\[ R = \frac{n^2}{Z} R_0 \]
where \( R_0 = 0.529 \) Å (angstroms).
- The velocity \( V \) of the electron in the \( n^{th} \) orbit is given by:
\[ V = \frac{Z}{n} V_0 \]
where \( V_0 \) is a constant.

### Step 3: Calculate the time period
The time period \( T \) can be calculated as:
\[ T = \frac{\text{Circumference of orbit}}{\text{Velocity}} = \frac{2 \pi R}{V} \]
Substituting the expressions for \( R \) and \( V \):
\[ T = \frac{2 \pi \left( \frac{n^2}{Z} R_0 \right)}{\left( \frac{Z}{n} V_0 \right)} \]
Simplifying this expression:
\[ T = \frac{2 \pi n^2 R_0}{Z} \cdot \frac{n}{Z V_0} = \frac{2 \pi n^3 R_0}{Z^2 V_0} \]

### Step 4: Identify constants
From the expression we derived:
\[ T = \frac{2 \pi R_0}{V_0} n^3 \frac{1}{Z^2} \]
This means:
- \( T_0 = \frac{2 \pi R_0}{V_0} \)
- \( A = 3 \)
- \( b = 2 \)

### Step 5: Calculate \( T_0 \)
Given that \( R_0 = 0.529 \) Å and \( V_0 \) is a known constant (approximately \( 2.18 \times 10^6 \) m/s), we can substitute these values to find \( T_0 \):
\[ T_0 = \frac{2 \pi (0.529 \times 10^{-10} \text{ m})}{2.18 \times 10^6 \text{ m/s}} \approx 1.51 \times 10^{-16} \text{ seconds} \]

### Final Results
Thus, we have:
- \( T_0 \approx 1.51 \times 10^{-16} \) seconds
- \( A = 3 \)
- \( b = 2 \)

### Summary
The final values are:
- \( T_0 = 1.51 \times 10^{-16} \) seconds
- \( A = 3 \)
- \( b = 2 \)