In a mixture of `H- He^(+)` gas (`He+ `is singly ionized He atom), `H` atom and `He+` ions are excited to their respective first excited state. Subsequently `H` atoms transfer their total excitation energy to `He+` ions (by collisions) Assume that the bohr model of atom is exactly valid.
The wavelength of light emitted in the visible region by `He+` lons after collisions with `H` atoms is -

Options

A. `6.5 xx 10^(-7) m`
B. `5.6 xx 10^(-7)m`
C. `4.8 xx 10^(-7) m`
D. `4.0 xx 10^(-7) m`

Explanation

To solve the problem, we need to determine the wavelength of light emitted by the singly ionized helium ion (He⁺) after it receives excitation energy from hydrogen (H) atoms. We will use the Bohr model of the atom to find the transition energy and then calculate the corresponding wavelength.

### Step-by-Step Solution:

1. **Identify the Energy Levels**:
For the He⁺ ion, the energy levels can be calculated using the formula:
\[ E_n = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \]
where \( Z \) is the atomic number (for He⁺, \( Z = 2 \)) and \( n \) is the principal quantum number.

2. **Determine the Excitation Energy**:
The hydrogen atom is excited to its first excited state, which corresponds to \( n = 2 \). The energy of the first excited state of hydrogen is:
\[ E_H = -\frac{1^2 \cdot 13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]
The energy of the ground state of hydrogen is \( -13.6 \, \text{eV} \), so the energy difference (excitation energy) is:
\[ \Delta E_H = E_{n=2} - E_{n=1} = -3.4 \, \text{eV} - (-13.6 \, \text{eV}) = 10.2 \, \text{eV} \]

3. **Transfer of Energy to He⁺**:
When the hydrogen atom transfers its excitation energy to the He⁺ ion, we need to find the energy levels of He⁺ for the transition that will emit light in the visible region. The relevant transition in He⁺ is from \( n = 4 \) to \( n = 3 \).

4. **Calculate the Energy for He⁺ Transition**:
The energy for the transition from \( n = 4 \) to \( n = 3 \) in He⁺ is given by:
\[ E_{He^+} = E_4 - E_3 = \left(-\frac{2^2 \cdot 13.6}{4^2}\right) - \left(-\frac{2^2 \cdot 13.6}{3^2}\right) \]
\[ E_{He^+} = -\frac{2^2 \cdot 13.6}{16} + \frac{2^2 \cdot 13.6}{9} \]
\[ = 13.6 \cdot 4 \left(\frac{1}{9} - \frac{1}{16}\right) \]
\[ = 13.6 \cdot 4 \left(\frac{16 - 9}{144}\right) = 13.6 \cdot 4 \cdot \frac{7}{144} \]
\[ = \frac{13.6 \cdot 28}{144} \approx 2.62 \, \text{eV} \]

5. **Calculate the Wavelength**:
The wavelength \( \lambda \) can be calculated using the energy-wavelength relation:
\[ E = \frac{hc}{\lambda} \]
Rearranging gives:
\[ \lambda = \frac{hc}{E} \]
Substituting \( h = 4.14 \times 10^{-15} \, \text{eV s} \) and \( c = 3 \times 10^8 \, \text{m/s} \):
\[ \lambda = \frac{(4.14 \times 10^{-15} \, \text{eV s})(3 \times 10^8 \, \text{m/s})}{2.62 \, \text{eV}} \]
\[ \lambda \approx \frac{1.242 \times 10^{-6}}{2.62} \approx 4.73 \times 10^{-7} \, \text{m} = 473 \, \text{nm} \]

### Final Answer:
The wavelength of light emitted in the visible region by He⁺ ions after collisions with H atoms is approximately \( 473 \, \text{nm} \).