A proton and an electron initially at rest are accelerated by the same potential difference. Assuming that a proton is 2000 times heavier than an electron, what will be the relation between the de Broglie wavelength of the proton `(lambda_p)` and that of electron `(lambda_e)`

Options

A. `lambda_p = 2000 lambda_e`
B. `lambda_p = (lambda_e)/(2000)`
C. `lambda_p = 20 sqrt(5) lambda_e`
D. `lambda_p = (lambda_e)/(20 sqrt(5))`

Explanation

To find the relation between the de Broglie wavelength of a proton \((\lambda_p)\) and that of an electron \((\lambda_e)\) when both are accelerated by the same potential difference, we can follow these steps:

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength \(\lambda\) 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.

### Step 2: Relate momentum to kinetic energy
When a charged particle is accelerated through a potential difference \(V\), its kinetic energy \(K\) can be expressed as:
\[ K = qV \]
where \(q\) is the charge of the particle. For the electron, \(q = e\) and for the proton, \(q = e\) as well (both have the same magnitude of charge).

### Step 3: Write the kinetic energy for both particles
The kinetic energy for both the electron and the proton can be expressed as:
\[ K_e = eV \quad \text{(for electron)} \]
\[ K_p = eV \quad \text{(for proton)} \]

### Step 4: Express momentum in terms of kinetic energy
The momentum \(p\) of a particle can be related to its kinetic energy \(K\) and mass \(m\) as follows:
\[ p = \sqrt{2mK} \]
Thus, for the electron and proton, we have:
\[ p_e = \sqrt{2m_e K_e} = \sqrt{2m_e eV} \]
\[ p_p = \sqrt{2m_p K_p} = \sqrt{2m_p eV} \]

### Step 5: Substitute momentum into the de Broglie wavelength formula
Now we can substitute these expressions for momentum into the de Broglie wavelength formula:
\[ \lambda_e = \frac{h}{p_e} = \frac{h}{\sqrt{2m_e eV}} \]
\[ \lambda_p = \frac{h}{p_p} = \frac{h}{\sqrt{2m_p eV}} \]

### Step 6: Find the ratio of the wavelengths
Now, we can find the ratio of the de Broglie wavelengths:
\[ \frac{\lambda_p}{\lambda_e} = \frac{\frac{h}{\sqrt{2m_p eV}}}{\frac{h}{\sqrt{2m_e eV}}} = \frac{\sqrt{2m_e eV}}{\sqrt{2m_p eV}} = \frac{\sqrt{m_e}}{\sqrt{m_p}} \]

### Step 7: Substitute the mass relationship
Given that the mass of the proton \(m_p\) is 2000 times the mass of the electron \(m_e\):
\[ m_p = 2000 m_e \]
Substituting this into the ratio gives:
\[ \frac{\lambda_p}{\lambda_e} = \frac{\sqrt{m_e}}{\sqrt{2000 m_e}} = \frac{\sqrt{m_e}}{\sqrt{2000} \sqrt{m_e}} = \frac{1}{\sqrt{2000}} = \frac{1}{20\sqrt{5}} \]

### Step 8: Final relation
Thus, we can express \(\lambda_p\) in terms of \(\lambda_e\):
\[ \lambda_p = \frac{\lambda_e}{20\sqrt{5}} \]

### Conclusion
The relation between the de Broglie wavelength of the proton and that of the electron is:
\[ \lambda_p = \frac{\lambda_e}{20\sqrt{5}} \]