An atom has a nucleus of charge +Ze, where Z is constant and e is electronic charge. An electron moving in a stationary orbit around nucleus requires an energy of 30.22 eV to excite from n=2 to n=3 energy state. Answer the following questions. Take ionisation energy of hydrogen as 13.6 eV.
Magnitude of angular momentum of electron in ground state is

Options

A. `2.01xx10^(-34) kg m^2 s^(-1)`
B. `1.05xx10^(-34) kg m^2 s^(-1)`
C. `1.575xx10^(-34) kg m^2 s^(-1)`
D. `3.201xx10^(-34) kg m^2 s^(-1)`

Explanation

To find the magnitude of the angular momentum of the electron in the ground state of the atom, we can use Bohr's second postulate of quantization of angular momentum. Here are the steps to derive the solution:

### Step 1: Understand the formula for angular momentum
According to Bohr's model, the angular momentum (L) of an electron in a stationary orbit is given by the formula:
\[ L = n \frac{h}{2\pi} \]
where:
- \( L \) is the angular momentum,
- \( n \) is the principal quantum number (for ground state, \( n = 1 \)),
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)).

### Step 2: Substitute the values
For the ground state, we have:
- \( n = 1 \)
- \( h = 6.626 \times 10^{-34} \, \text{Js} \)

Substituting these values into the formula:
\[ L = 1 \cdot \frac{6.626 \times 10^{-34}}{2\pi} \]

### Step 3: Calculate \( 2\pi \)
Calculate \( 2\pi \):
\[ 2\pi \approx 6.2832 \]

### Step 4: Perform the division
Now, perform the division:
\[ L = \frac{6.626 \times 10^{-34}}{6.2832} \]

Calculating this gives:
\[ L \approx 1.055 \times 10^{-34} \, \text{kg m}^2/\text{s} \]

### Step 5: Round to significant figures
Rounding to two decimal places, we find:
\[ L \approx 1.05 \times 10^{-34} \, \text{kg m}^2/\text{s} \]

### Final Answer
Thus, the magnitude of the angular momentum of the electron in the ground state is:
\[ \boxed{1.05 \times 10^{-34} \, \text{kg m}^2/\text{s}} \]

---