Electron are accelerated through a potential difference V and protons are accelerated through a potential difference of 4 V. The de-Broglie wavelength are `lamda_e and lamda_p` for electrons and protons, respectively The ratio of `lamda_e/lamda_p` is given by ( given , `m_e` is mass of electron and `m_p` is mass of proton )

Options

A. `lamda_e/lamda_p=sqrt((m_p)/m_e)`
B. `lamda_e/lamda_p=sqrt((m_e)/m_p)`
C. `lamda_e/lamda_p=1/2sqrt((m_e)/m_p)`
D. `lamda_e/lamda_p=2sqrt((m_p)/m_e)`

Answer

To solve the problem, we need to find the ratio of the de-Broglie wavelengths of electrons and protons when they are accelerated through different potential differences.

### Step-by-Step Solution:

1. **Understanding de-Broglie Wavelength**:
The de-Broglie wavelength (λ) of a particle is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

2. **Relating Momentum to Kinetic Energy**:
The kinetic energy (KE) of a charged particle accelerated through a potential difference \( V \) is given by:
\[ KE = QV \]
where \( Q \) is the charge of the particle. For an electron (charge \( e \)), the kinetic energy becomes:
\[ KE_e = eV \]
For a proton (charge \( e \)), when accelerated through a potential difference of \( 4V \), the kinetic energy is:
\[ KE_p = e(4V) = 4eV \]

3. **Finding Momentum**:
The momentum \( p \) of a particle can be expressed in terms of its kinetic energy:
\[ p = \sqrt{2m \cdot KE} \]
Therefore, for the electron:
\[ p_e = \sqrt{2m_e \cdot eV} \]
And for the proton:
\[ p_p = \sqrt{2m_p \cdot 4eV} = \sqrt{8m_p \cdot eV} \]

4. **Calculating de-Broglie Wavelengths**:
Now we can express the de-Broglie wavelengths for the electron and proton:
\[ \lambda_e = \frac{h}{p_e} = \frac{h}{\sqrt{2m_e \cdot eV}} \]
\[ \lambda_p = \frac{h}{p_p} = \frac{h}{\sqrt{8m_p \cdot eV}} \]

5. **Finding the Ratio**:
To find the ratio \( \frac{\lambda_e}{\lambda_p} \):
\[ \frac{\lambda_e}{\lambda_p} = \frac{\frac{h}{\sqrt{2m_e \cdot eV}}}{\frac{h}{\sqrt{8m_p \cdot eV}}} = \frac{\sqrt{8m_p \cdot eV}}{\sqrt{2m_e \cdot eV}} \]
The \( h \) and \( eV \) cancel out:
\[ = \frac{\sqrt{8m_p}}{\sqrt{2m_e}} = \frac{\sqrt{8}}{\sqrt{2}} \cdot \sqrt{\frac{m_p}{m_e}} = 2\sqrt{2} \cdot \sqrt{\frac{m_p}{m_e}} \]

6. **Final Result**:
Thus, the ratio of the de-Broglie wavelengths is:
\[ \frac{\lambda_e}{\lambda_p} = 2\sqrt{2} \cdot \sqrt{\frac{m_p}{m_e}} \]