Two electrons are moving with the same speed v. One electron enters a region of uniform electric field while the other enters a region of uniform magnetic field,then after sometime if the de-Broglie wavelenths of the two are `lambda_(1)` and `lambda_(2)` then `:`

Options

A. `lambda_(1) =lambda_(2)`
B. `lambda_(1) gt lambda_(2)`
C. `lambda_(1) lt lambda_(2)`
D. `lambda_(1) gt lambda_(2)` or `lambda_(1) lt lambda_(2)`

Answer

To solve the problem, we need to analyze the behavior of two electrons moving with the same initial speed \( v \) when they enter different fields: one enters an electric field and the other enters a magnetic field. We will derive the de-Broglie wavelengths \( \lambda_1 \) and \( \lambda_2 \) for each case and compare them.

### Step-by-Step Solution:

1. **Initial de-Broglie Wavelength Calculation**:
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.

For both electrons before entering the fields, their initial de-Broglie wavelengths are:
\[ \lambda_0 = \frac{h}{mv} \]

2. **Electron in Electric Field**:
When the first electron enters a uniform electric field \( E \), it experiences a force \( F = eE \) (where \( e \) is the charge of the electron). This force causes the electron to accelerate:
\[ F = ma \implies a = \frac{eE}{m} \]
The electron will gain velocity due to this acceleration. If we denote the final velocity after some time as \( v_T \), we can say:
\[ v_T = v + at = v + \frac{eE}{m}t \]
The de-Broglie wavelength for the electron in the electric field becomes:
\[ \lambda_1 = \frac{h}{m v_T} = \frac{h}{m \left(v + \frac{eE}{m}t\right)} \]

3. **Electron in Magnetic Field**:
The second electron enters a uniform magnetic field \( B \). The magnetic force acts as a centripetal force, causing the electron to move in a circular path. The magnetic force is given by:
\[ F = evB \]
This force provides the necessary centripetal force:
\[ evB = \frac{mv^2}{r} \implies r = \frac{mv}{eB} \]
The speed of the electron remains constant in a magnetic field (it does not gain or lose speed, only changes direction), so its de-Broglie wavelength remains:
\[ \lambda_2 = \frac{h}{mv} \]

4. **Comparison of Wavelengths**:
Now we compare \( \lambda_1 \) and \( \lambda_2 \):
- Since \( v_T > v \) for the electron in the electric field, we have:
\[ \lambda_1 = \frac{h}{m \left(v + \frac{eE}{m}t\right)} < \frac{h}{mv} = \lambda_2 \]
- Therefore, we conclude that:
\[ \lambda_1 < \lambda_2 \]

### Final Result:
The relationship between the de-Broglie wavelengths is:
\[ \lambda_1 < \lambda_2 \]