Consider `3rd` orbit of `He^(+)` (Helium) using nonrelativistic approach the speed of electron in this orbit will be (given `K = 9 xx 10^(9)` constant `Z = 2` and `h` (Planck's constant) `= 6.6 xx 10^(-34)Js`.)

Options

A. `2.92xx10^(6)m//s`
B. `1.46xx10^(6)m//s`
C. `0.73xx10^(6)m//s`
D. `3.0xx10^(8)m//s`

Answer

To find the speed of the electron in the third orbit of the Helium ion (He⁺), we can follow these steps:

### Step 1: Determine the Energy in the Third Orbit
The energy of an electron in the nth orbit of a hydrogen-like atom is given by the formula:

\[ E_n = -\frac{13.6 Z^2}{n^2} \text{ eV} \]

Where:
- \( Z \) is the atomic number (for He⁺, \( Z = 2 \))
- \( n \) is the principal quantum number (for the third orbit, \( n = 3 \))

Substituting the values:

\[ E_3 = -\frac{13.6 \times 2^2}{3^2} = -\frac{13.6 \times 4}{9} = -\frac{54.4}{9} \approx -6.04 \text{ eV} \]

### Step 2: Convert Energy to Joules
To convert the energy from electron volts to joules, we use the conversion factor \( 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \):

\[ E_3 = -6.04 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} \approx -9.664 \times 10^{-19} \text{ J} \]

### Step 3: Relate Energy to Kinetic Energy
In a hydrogen-like atom, the kinetic energy (K.E.) of the electron is equal to half the magnitude of the total energy (since potential energy is twice the kinetic energy in a bound system):

\[ K.E. = -\frac{E_3}{2} = \frac{9.664 \times 10^{-19}}{2} \approx 4.832 \times 10^{-19} \text{ J} \]

### Step 4: Use Kinetic Energy to Find Speed
The kinetic energy is also given by the formula:

\[ K.E. = \frac{1}{2} m v^2 \]

Where \( m \) is the mass of the electron (\( m \approx 9.1 \times 10^{-31} \text{ kg} \)) and \( v \) is the speed of the electron. Rearranging the formula to solve for \( v \):

\[ v^2 = \frac{2 \times K.E.}{m} = \frac{2 \times 4.832 \times 10^{-19}}{9.1 \times 10^{-31}} \]

Calculating \( v^2 \):

\[ v^2 \approx \frac{9.664 \times 10^{-19}}{9.1 \times 10^{-31}} \approx 1.06 \times 10^{12} \]

### Step 5: Calculate the Speed
Taking the square root to find \( v \):

\[ v \approx \sqrt{1.06 \times 10^{12}} \approx 1.03 \times 10^6 \text{ m/s} \]

### Final Answer
Thus, the speed of the electron in the third orbit of He⁺ is approximately:

\[ v \approx 1.03 \times 10^6 \text{ m/s} \]