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. `2.29xx10^(6)ms^(-1)`
B. `1.46xx10^(6)ms^(-1)`
C. `0.73xx10^(6)ms^(-1)`
D. `3xx10^(8)ms^(-1)`

Explanation

To find the speed of the electron in the third orbit of the helium ion \( \text{He}^+ \) using a non-relativistic approach, we can follow these steps:

### Step 1: Calculate the energy of the electron in the third orbit
The energy of the 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 \( \text{He}^+ \), \( Z = 2 \))
- \( n \) is the principal quantum number (for the third orbit, \( n = 3 \))

Substituting the values:

\[ E_3 = -\frac{13.6 \times 2^2}{3^2} \, \text{eV} \]

Calculating this:

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

To convert this energy into joules, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):

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

### Step 2: Relate kinetic energy to the energy of the electron
According to Bohr's model, the kinetic energy \( KE \) of the electron is given by:

\[ KE = -E_n \]

Thus, we have:

\[ KE = 9.664 \times 10^{-19} \, \text{J} \]

### Step 3: Use the kinetic energy formula to find speed
The kinetic energy can also be expressed in terms of mass and speed:

\[ KE = \frac{1}{2} m_e v^2 \]

Where \( m_e \) is the mass of the electron, approximately \( 9.1 \times 10^{-31} \, \text{kg} \).

Setting the two expressions for kinetic energy equal gives:

\[ 9.664 \times 10^{-19} = \frac{1}{2} (9.1 \times 10^{-31}) v^2 \]

### Step 4: Solve for \( v \)
Rearranging the equation to solve for \( v^2 \):

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

Calculating this:

\[ v^2 = \frac{1.9328 \times 10^{-18}}{9.1 \times 10^{-31}} \approx 2.12 \times 10^{12} \]

Taking the square root to find \( v \):

\[ v \approx \sqrt{2.12 \times 10^{12}} \approx 1.46 \times 10^6 \, \text{m/s} \]

### Final Answer
The speed of the electron in the third orbit of \( \text{He}^+ \) is approximately:

\[ v \approx 1.46 \times 10^6 \, \text{m/s} \]

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