An electron accelerated under a potential difference of V volt has a certain wavelength `lamda` It is known that the mass of a proton is about 1800 times the mass of an electron . If a proton has to have the same wavelength `lamda`, then it will have to be accelerated under the potential difference of

Options

A. V volt
B. 1800 volt
C. `V/1800 `volt
D. `sqrt1800` volt

Answer

To solve the problem step by step, we will use the de Broglie wavelength formula and the relationship between kinetic energy and potential difference.

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength (λ) of a particle is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is the Planck constant and \( p \) is the momentum of the particle.

### Step 2: Relate momentum to kinetic energy
The momentum \( p \) can be expressed in terms of kinetic energy (K.E.):
\[ K.E. = \frac{p^2}{2m} \implies p = \sqrt{2m \cdot K.E.} \]

### Step 3: Relate kinetic energy to potential difference
When a charged particle is accelerated through a potential difference \( V \), the kinetic energy gained is given by:
\[ K.E. = qV \]
where \( q \) is the charge of the particle.

### Step 4: Substitute for momentum in terms of potential difference
For an electron, the kinetic energy can be expressed as:
\[ K.E. = eV \quad \text{(where \( e \) is the charge of the electron)} \]
Thus, the momentum of the electron becomes:
\[ p_e = \sqrt{2m_e \cdot eV} \]

### Step 5: Substitute into the de Broglie wavelength formula
Substituting this expression for momentum into the de Broglie wavelength formula for the electron gives:
\[ \lambda = \frac{h}{\sqrt{2m_e \cdot eV}} \]

### Step 6: Set up the equation for the proton
For the proton to have the same wavelength \( \lambda \), we can set up a similar equation:
\[ \lambda = \frac{h}{\sqrt{2m_p \cdot qV'}} \]
where \( V' \) is the potential difference required for the proton and \( m_p \) is the mass of the proton.

### Step 7: Equate the two expressions for wavelength
Since both expressions equal \( \lambda \), we can set them equal to each other:
\[ \frac{h}{\sqrt{2m_e \cdot eV}} = \frac{h}{\sqrt{2m_p \cdot qV'}} \]

### Step 8: Cancel \( h \) and rearrange
Cancelling \( h \) from both sides and rearranging gives:
\[ \sqrt{2m_e \cdot eV} = \sqrt{2m_p \cdot qV'} \]

### Step 9: Square both sides
Squaring both sides leads to:
\[ 2m_e \cdot eV = 2m_p \cdot qV' \]

### Step 10: Substitute \( m_p = 1800m_e \)
Substituting \( m_p = 1800m_e \) into the equation gives:
\[ 2m_e \cdot eV = 2(1800m_e) \cdot qV' \]

### Step 11: Simplify the equation
We can cancel \( 2m_e \) from both sides:
\[ eV = 1800qV' \]

### Step 12: Solve for \( V' \)
Rearranging gives:
\[ V' = \frac{eV}{1800q} \]
Since for both the electron and proton, \( q \) is the same (charge), we can simplify further:
\[ V' = \frac{V}{1800} \]

### Conclusion
Thus, the potential difference \( V' \) that the proton must be accelerated under to have the same wavelength \( \lambda \) as the electron is:
\[ V' = \frac{V}{1800} \]