An electron and a proton are accelerated by same voltage difference. Find the ratio of the de broglie wavelength of electron and proton

Options

A. `1860/1`
B. `41.4/1`
C. `43/1`
D. `4/1`

Answer

To find the ratio of the de Broglie wavelengths of an electron and a proton when both are accelerated by the same voltage 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
The momentum \( p \) of a particle can also be expressed in terms of its mass \( m \) and velocity \( v \):
\[ p = mv \]
The kinetic energy (KE) of a particle is given by:
\[ KE = \frac{1}{2} mv^2 \]

### Step 3: Express kinetic energy in terms of voltage
When a charged particle (like an electron or a proton) is accelerated through a voltage \( V \), the kinetic energy gained by the particle is equal to the work done on it, which can be expressed as:
\[ KE = qV \]
where \( q \) is the charge of the particle. For both the electron and proton, the charge \( q \) is the same in magnitude but different in sign.

### Step 4: Set up the equations for both particles
For an electron:
\[ KE_e = eV \]
For a proton:
\[ KE_p = eV \]
where \( e \) is the elementary charge.

### Step 5: Relate the momentum to kinetic energy
From the kinetic energy expressions, we can express the momentum in terms of kinetic energy:
\[ p = \sqrt{2m \cdot KE} \]
Thus for the electron:
\[ p_e = \sqrt{2m_e \cdot eV} \]
And for the proton:
\[ p_p = \sqrt{2m_p \cdot eV} \]

### Step 6: Substitute momentum into the de Broglie wavelength formula
Now substituting the momentum expressions into the de Broglie wavelength formula:
\[ \lambda_e = \frac{h}{p_e} = \frac{h}{\sqrt{2m_e \cdot eV}} \]
\[ \lambda_p = \frac{h}{p_p} = \frac{h}{\sqrt{2m_p \cdot eV}} \]

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

### Step 8: Substitute the masses of the electron and proton
The mass of the proton \( m_p \) is approximately \( 1.67 \times 10^{-27} \) kg and the mass of the electron \( m_e \) is approximately \( 9.11 \times 10^{-31} \) kg. Thus:
\[ \frac{\lambda_e}{\lambda_p} = \sqrt{\frac{1.67 \times 10^{-27}}{9.11 \times 10^{-31}}} \]

### Step 9: Calculate the ratio
Calculating the above expression gives:
\[ \frac{\lambda_e}{\lambda_p} \approx \sqrt{1836} \approx 43 \]

### Final Answer
The ratio of the de Broglie wavelength of an electron to that of a proton is approximately:
\[ \frac{\lambda_e}{\lambda_p} \approx 43 \]