A photon of wavelength `0.1 A` is emitted by a helium atom as a consequence of the emission of photon. The KE gained by helium atom is

Options

A. 0.05eV
B. 1.05 eV
C. 2.05 eV
D. 3.05 eV

Explanation

To solve the problem of finding the kinetic energy gained by a helium atom when a photon of wavelength \(0.1 \, \text{Å}\) is emitted, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between energy and momentum**:
The kinetic energy (KE) gained by the helium atom can be expressed in terms of momentum \(p\):
\[ KE = \frac{p^2}{2m} \]
where \(m\) is the mass of the helium atom.

2. **Relate momentum to wavelength**:
The momentum \(p\) of a photon can be expressed using the Planck's constant \(h\) and the wavelength \(\lambda\):
\[ p = \frac{h}{\lambda} \]

3. **Substitute momentum into the kinetic energy formula**:
Substitute the expression for momentum into the kinetic energy formula:
\[ KE = \frac{(h/\lambda)^2}{2m} = \frac{h^2}{2m\lambda^2} \]

4. **Insert known values**:
- Planck's constant \(h = 6.63 \times 10^{-34} \, \text{Js}\)
- Mass of helium atom \(m = 4 \times 1.67 \times 10^{-27} \, \text{kg} = 6.68 \times 10^{-27} \, \text{kg}\)
- Wavelength \(\lambda = 0.1 \, \text{Å} = 0.1 \times 10^{-10} \, \text{m} = 1 \times 10^{-11} \, \text{m}\)

5. **Calculate the kinetic energy**:
Substitute the values into the kinetic energy formula:
\[ KE = \frac{(6.63 \times 10^{-34})^2}{2 \times (6.68 \times 10^{-27}) \times (1 \times 10^{-11})^2} \]

Calculating the numerator:
\[ (6.63 \times 10^{-34})^2 = 4.39 \times 10^{-67} \]

Calculating the denominator:
\[ 2 \times (6.68 \times 10^{-27}) \times (1 \times 10^{-11})^2 = 2 \times 6.68 \times 10^{-27} \times 1 \times 10^{-22} = 1.336 \times 10^{-48} \]

Now, substituting these into the equation:
\[ KE = \frac{4.39 \times 10^{-67}}{1.336 \times 10^{-48}} \approx 3.29 \times 10^{-19} \, \text{J} \]

6. **Convert the energy to electron volts**:
To convert joules to electron volts, use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):
\[ KE \approx \frac{3.29 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.06 \, \text{eV} \]

### Final Answer:
The kinetic energy gained by the helium atom is approximately \(2.06 \, \text{eV}\).

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