In 1924, de-Broglie proposed that every particle possesses wave properties with a wavelength , `lambda` given by `lambda = h/(mv)` where m is the mass of the particle, `v` is its velocity and h is Planck's constant. The de-Broglie prediction was confirmed experimentally when it was found that an electron beam undergoes diffraction , a phenomenon characteristic of waves. The de-Broglie wavelength can be estimated by measuring kinetic energy of an electron accelerating by a potential V as :
`1/2 mv^2 = eV "where" 1eV = 1.6 xx 10^(-19) J, h = 6.6 xx 10^(-34) Js`.
The proton and `He^(2+)` are accelerated by the same potential, then their de-Broglie wavelengths `lambda_(He^(2+)) and lambda_p` are in the ratio of `(m_(He^(2+)) = 4m_p)` :

Options

A. `1/(sqrt2)`
B. `1/(2sqrt(2))`
C. `2sqrt(2)`
D. `1/2`

Explanation

To solve the problem, we need to find the ratio of the de Broglie wavelengths of a proton and a Helium ion (He²⁺) when both are accelerated by the same potential. We will use the de Broglie wavelength formula and the relationship between kinetic energy and potential energy.

### Step-by-Step Solution:

1. **Understanding Kinetic Energy**:
The kinetic energy (KE) of a particle accelerated by a potential \( V \) is given by:
\[ KE = eV \]
where \( e \) is the charge of the particle.

2. **Kinetic Energy Expression**:
For a particle of mass \( m \) and velocity \( v \), the kinetic energy can also be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
Equating the two expressions for kinetic energy, we have:
\[ \frac{1}{2} mv^2 = eV \]

3. **Solving for Velocity**:
Rearranging the equation to solve for \( v \):
\[ mv^2 = 2eV \implies v^2 = \frac{2eV}{m} \implies v = \sqrt{\frac{2eV}{m}} \]

4. **Finding the de Broglie Wavelength**:
The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{mv} \]
Substituting \( v \) from the previous step:
\[ \lambda = \frac{h}{m \sqrt{\frac{2eV}{m}}} = \frac{h \sqrt{m}}{\sqrt{2eV}} \]

5. **Calculating Wavelengths for Proton and Helium Ion**:
- For the proton (mass \( m_p \)):
\[ \lambda_p = \frac{h \sqrt{m_p}}{\sqrt{2eV}} \]
- For the Helium ion (mass \( m_{He^{2+}} = 4m_p \)):
\[ \lambda_{He^{2+}} = \frac{h \sqrt{4m_p}}{\sqrt{2eV}} = \frac{2h \sqrt{m_p}}{\sqrt{2eV}} \]

6. **Finding the Ratio of Wavelengths**:
Now we find the ratio of the wavelengths:
\[ \frac{\lambda_{He^{2+}}}{\lambda_p} = \frac{2h \sqrt{m_p}/\sqrt{2eV}}{h \sqrt{m_p}/\sqrt{2eV}} = \frac{2}{1} = 2 \]

7. **Final Ratio**:
To find the ratio of the de Broglie wavelengths:
\[ \frac{\lambda_{He^{2+}}}{\lambda_p} = \frac{1}{\sqrt{4}} = \frac{1}{2} \]

### Conclusion:
The ratio of the de Broglie wavelengths of the Helium ion and the proton is:
\[ \frac{\lambda_{He^{2+}}}{\lambda_p} = \frac{1}{2} \]