The radius of electron orbit and the speed of electron in the ground state of hydrogen atom is `5.30xx10^(-11)m` and `22xx10^(6)m s^(-1)` respectively, then the orbital period of this electron in second excited state will be

Options

A. `1.21xx10^(-14)s`
B. `1.21xx10^(-12)s`
C. `1.21xx10^(-10)s`
D. `1.21xx10^(-15)s`

Explanation

To find the orbital period of the electron in the second excited state of a hydrogen atom, we can follow these steps:

### Step 1: Understand the relationship for radius and speed
The radius of the electron's orbit in the nth state of a hydrogen atom is given by:
\[ r_n = n^2 \cdot r_0 \]
where \( r_0 \) is the radius of the ground state (first orbit), which is given as \( 5.30 \times 10^{-11} \, m \).

### Step 2: Calculate the radius for the second excited state (n=3)
The second excited state corresponds to \( n = 3 \):
\[ r_3 = 3^2 \cdot r_0 = 9 \cdot (5.30 \times 10^{-11}) \]
\[ r_3 = 4.77 \times 10^{-10} \, m \]

### Step 3: Understand the relationship for speed
The speed of the electron in the nth orbit is inversely proportional to n:
\[ v_n = \frac{v_1}{n} \]
where \( v_1 \) is the speed in the ground state, given as \( 22 \times 10^6 \, m/s \).

### Step 4: Calculate the speed for the second excited state (n=3)
\[ v_3 = \frac{22 \times 10^6}{3} \]
\[ v_3 = 7.33 \times 10^6 \, m/s \]

### Step 5: Calculate the orbital period
The orbital period \( T_n \) can be calculated using the formula:
\[ T_n = \frac{2 \pi r_n}{v_n} \]

Substituting the values we calculated:
\[ T_3 = \frac{2 \pi (4.77 \times 10^{-10})}{7.33 \times 10^6} \]

### Step 6: Perform the calculation
Calculating the numerator:
\[ 2 \pi (4.77 \times 10^{-10}) \approx 3.00 \times 10^{-9} \]

Now, calculate the period:
\[ T_3 = \frac{3.00 \times 10^{-9}}{7.33 \times 10^6} \]
\[ T_3 \approx 4.09 \times 10^{-16} \, s \]

### Final Answer
The orbital period of the electron in the second excited state is approximately:
\[ T_3 \approx 4.09 \times 10^{-16} \, s \]

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