What possibly can be the ratio of the de Broglie wavelength for two electrons each having zero initial energy and accelerated through 50 volts and 200 volts?

Options

A. `3:10`
B. `10:3`
C. `1:2`
D. `2:1`

Explanation

To find the ratio of the de Broglie wavelengths of two electrons accelerated through different voltages, 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:
\[ \lambda = \frac{h}{mv} \]
where \( h \) is Planck's constant, \( m \) is the mass of the electron, and \( v \) is its velocity.

### Step 2: Relate kinetic energy to voltage
When an electron is accelerated through a potential difference \( V \), it gains kinetic energy equal to the work done on it by the electric field:
\[ KE = eV \]
where \( e \) is the charge of the electron. The kinetic energy can also be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
Setting these equal gives:
\[ eV = \frac{1}{2} mv^2 \]

### Step 3: Solve for velocity \( v \)
Rearranging the equation for \( v \):
\[ v = \sqrt{\frac{2eV}{m}} \]

### Step 4: Substitute \( v \) into the de Broglie wavelength formula
Substituting \( v \) into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{m \sqrt{\frac{2eV}{m}}} = \frac{h}{\sqrt{2emV}} \]

### Step 5: Determine the ratio of wavelengths for different voltages
Let \( V_1 = 50 \) volts and \( V_2 = 200 \) volts. The wavelengths for these voltages are:
\[ \lambda_1 = \frac{h}{\sqrt{2emV_1}} \quad \text{and} \quad \lambda_2 = \frac{h}{\sqrt{2emV_2}} \]
Now, we can find the ratio:
\[ \frac{\lambda_1}{\lambda_2} = \frac{\sqrt{2emV_2}}{\sqrt{2emV_1}} = \sqrt{\frac{V_2}{V_1}} \]

### Step 6: Substitute the values of \( V_1 \) and \( V_2 \)
Substituting \( V_1 = 50 \) volts and \( V_2 = 200 \) volts:
\[ \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{200}{50}} = \sqrt{4} = 2 \]

### Conclusion
Thus, the ratio of the de Broglie wavelengths of the two electrons is:
\[ \frac{\lambda_1}{\lambda_2} = 2:1 \]