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 solve the problem of finding the time period of revolution of an electron in the nth orbit of a hydrogen-like atom, we can follow these steps:

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

### Step 2: Determine the radius of the orbit
According to Bohr's model, the radius \( r \) of the nth orbit in a hydrogen-like atom is given by:
\[ r = \frac{n^2}{Z} r_0 \]
where \( r_0 \) is the Bohr radius, approximately \( 0.529 \) Å.

### Step 3: Determine the speed of the electron
The speed \( v \) of the electron in the nth orbit is given by:
\[ v = \frac{Z}{n} v_0 \]
where \( v_0 \) is the speed of the electron in the first orbit, approximately \( 2.2 \times 10^8 \) m/s.

### Step 4: Calculate the time period
The time period \( T \) can be calculated using the formula:
\[ T = \frac{2\pi r}{v} \]
Substituting the expressions for \( r \) and \( v \):
\[ T = \frac{2\pi \left(\frac{n^2}{Z} r_0\right)}{\left(\frac{Z}{n} v_0\right)} \]
This simplifies to:
\[ T = \frac{2\pi n^3 r_0}{Z^2 v_0} \]

### Step 5: Identify constants
From the derived formula, we can identify:
- \( T_0 = \frac{2\pi r_0}{v_0} \)
- \( a = 3 \)
- \( b = 2 \)

### Step 6: Calculate \( T_0 \)
Substituting the known values:
- \( r_0 = 0.529 \times 10^{-10} \) m
- \( v_0 = 2.2 \times 10^8 \) m/s

Calculating \( T_0 \):
\[ T_0 = \frac{2\pi (0.529 \times 10^{-10})}{2.2 \times 10^8} \]
Calculating this gives:
\[ T_0 \approx 1.51 \times 10^{-16} \text{ seconds} \]

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

### Conclusion
The values are:
- \( T_0 = 1.51 \times 10^{-16} \)
- \( a = 3 \)
- \( b = 2 \)