An electron and a proton are accelerated through the same potential difference. The ratio of their de broglic wavelengths will be

Options

A. `(m_(p))/(m_(e ))`
B. `(M_(p)v_(e ))/(m_(e )v_(p))`
C. `(m_(p)^(2))/(m_(e )^(2))`
D. `sqrt(m_(p))/(m_(e ))`

Answer

To find the ratio of the de Broglie wavelengths of an electron and a proton when both are accelerated through the same potential difference, we can follow these steps:

### Step 1: Understand the relationship between kinetic energy and potential difference
When a charged particle is accelerated through a potential difference (V), its kinetic energy (KE) can be expressed as:
\[ KE = Q \times V \]
where \( Q \) is the charge of the particle. For both the electron and proton, the charge is the same in magnitude but opposite in sign, so we can denote the charge as \( e \).

### Step 2: Write the kinetic energy for both particles
For the electron:
\[ KE_e = eV \]
For the proton:
\[ KE_p = eV \]
Since both particles are accelerated through the same potential difference, their kinetic energies are equal.

### Step 3: Relate kinetic energy to momentum
The kinetic energy can also be expressed in terms of momentum (p):
\[ KE = \frac{p^2}{2m} \]
where \( m \) is the mass of the particle. Rearranging this gives:
\[ p = \sqrt{2m \cdot KE} \]

### Step 4: Substitute the kinetic energy into the momentum equation
For the electron:
\[ p_e = \sqrt{2m_e \cdot eV} \]
For the proton:
\[ p_p = \sqrt{2m_p \cdot eV} \]

### Step 5: Write the de Broglie wavelength
The de Broglie wavelength (\( \lambda \)) is given by:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant.

### Step 6: Write the de Broglie wavelengths for both particles
For the electron:
\[ \lambda_e = \frac{h}{p_e} = \frac{h}{\sqrt{2m_e \cdot eV}} \]
For the proton:
\[ \lambda_p = \frac{h}{p_p} = \frac{h}{\sqrt{2m_p \cdot eV}} \]

### Step 7: Find the ratio of the de Broglie wavelengths
Now, we can find the ratio of the de Broglie wavelengths:
\[ \frac{\lambda_e}{\lambda_p} = \frac{\frac{h}{\sqrt{2m_e \cdot eV}}}{\frac{h}{\sqrt{2m_p \cdot eV}}} \]
This simplifies to:
\[ \frac{\lambda_e}{\lambda_p} = \frac{\sqrt{2m_p \cdot eV}}{\sqrt{2m_e \cdot eV}} = \sqrt{\frac{m_p}{m_e}} \]

### Step 8: Conclusion
Thus, the ratio of the de Broglie wavelengths of the electron and proton is:
\[ \frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}} \]