Speed of an electron Bohr's `7^( th)` orbit for Hydrogen atom is `3.6times10^(6)m/s`.The corresponding speed of the electron in `3^(rd)` orbit,in `m/s` is

Options

A. `3.6times10^(6)`
B. `7.5times10^(6)`
C. `1.8times10^(6)`
D. `8.4times10^(6)`

Answer

To find the speed of an electron in the 3rd orbit of a hydrogen atom given the speed in the 7th orbit, we can use the formula for the speed of an electron in the nth orbit of a hydrogen atom:

\[ v_n = \frac{Z e^2}{2 \epsilon_0 h} \cdot \frac{1}{n} \]

Where:
- \( v_n \) is the speed of the electron in the nth orbit,
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( e \) is the charge of the electron,
- \( \epsilon_0 \) is the permittivity of free space,
- \( h \) is Planck's constant,
- \( n \) is the principal quantum number (orbit number).

From the formula, we can see that the speed of the electron is inversely proportional to the principal quantum number \( n \). Therefore, we can express the relationship between the speeds in different orbits as:

\[ \frac{v_7}{v_3} = \frac{n_3}{n_7} \]

Where:
- \( v_7 \) is the speed in the 7th orbit,
- \( v_3 \) is the speed in the 3rd orbit,
- \( n_7 = 7 \) and \( n_3 = 3 \).

Now, we can rearrange this equation to find \( v_3 \):

\[ v_3 = v_7 \cdot \frac{n_7}{n_3} \]

Substituting the known values:

\[ v_3 = 3.6 \times 10^6 \, \text{m/s} \cdot \frac{7}{3} \]

Calculating \( v_3 \):

1. First, calculate \( \frac{7}{3} \):

\[ \frac{7}{3} \approx 2.3333 \]

2. Now multiply this by \( 3.6 \times 10^6 \):

\[ v_3 = 3.6 \times 10^6 \cdot 2.3333 \approx 8.4 \times 10^6 \, \text{m/s} \]

Thus, the speed of the electron in the 3rd orbit is approximately:

\[ v_3 \approx 8.4 \times 10^6 \, \text{m/s} \]

### Final Answer:
The corresponding speed of the electron in the 3rd orbit is \( 8.4 \times 10^6 \, \text{m/s} \).