An electron (mass m) with initial velocity `vecv = v_0 hati + 2v_0 hatj` is in an electric field `vecE = E_0 hatk` . If `lamda_0` is initial de-Broglie wavelength of electron, its de-Broglie wave length at time t is given by :

Options

A. `(lamda_0 sqrt2)/sqrt(1+(e^2 E^2t^2)/(m^2 v_0^2))`
B. `(lamda_0 sqrt2)/sqrt(2+(e^2 E^2t^2)/(m^2 v_0^2))`
C. `(lamda_0 sqrt2)/sqrt(3+(e^2 E^2t^2)/(m^2 v_0^2))`
D. `(lamda_0 )/sqrt(1+(e^2 E^2t^2)/(5m^2 v_0^2))`

Explanation

To solve the problem, we need to find the de-Broglie wavelength of the electron at time \( t \) given its initial conditions and the electric field it is subjected to. Let's break it down step by step:

### Step 1: Identify the initial velocity components
The initial velocity of the electron is given as:
\[ \vec{v} = v_0 \hat{i} + 2v_0 \hat{j} \]
This indicates that the initial velocity has components \( v_x = v_0 \) and \( v_y = 2v_0 \).

### Step 2: Determine the acceleration due to the electric field
The electron experiences an acceleration in the \( z \)-direction due to the electric field \( \vec{E} = E_0 \hat{k} \). The force on the electron is given by:
\[ \vec{F} = q \vec{E} \]
where \( q \) is the charge of the electron. The acceleration \( a_z \) in the \( z \)-direction can be calculated using Newton's second law:
\[ a_z = \frac{F}{m} = \frac{qE_0}{m} \]
Thus, the velocity in the \( z \)-direction after time \( t \) is:
\[ v_z = a_z t = \frac{qE_0}{m} t \]

### Step 3: Calculate the total velocity at time \( t \)
The total velocity \( \vec{v}_{\text{total}} \) at time \( t \) can be expressed as:
\[ \vec{v}_{\text{total}} = v_0 \hat{i} + 2v_0 \hat{j} + v_z \hat{k} \]
Substituting \( v_z \):
\[ \vec{v}_{\text{total}} = v_0 \hat{i} + 2v_0 \hat{j} + \left(\frac{qE_0}{m} t\right) \hat{k} \]

### Step 4: Calculate the magnitude of the total velocity
The magnitude of the total velocity \( v_{\text{total}} \) is given by:
\[ v_{\text{total}} = \sqrt{v_x^2 + v_y^2 + v_z^2} = \sqrt{(v_0)^2 + (2v_0)^2 + \left(\frac{qE_0}{m} t\right)^2} \]
This simplifies to:
\[ v_{\text{total}} = \sqrt{v_0^2 + 4v_0^2 + \left(\frac{qE_0}{m} t\right)^2} = \sqrt{5v_0^2 + \left(\frac{qE_0}{m} t\right)^2} \]

### Step 5: Relate the de-Broglie wavelength to the velocity
The de-Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{mv} \]
where \( h \) is Planck's constant and \( v \) is the velocity. The initial de-Broglie wavelength \( \lambda_0 \) is:
\[ \lambda_0 = \frac{h}{m v_0} \]
At time \( t \), the new de-Broglie wavelength \( \lambda_t \) is:
\[ \lambda_t = \frac{h}{m v_{\text{total}}} \]

### Step 6: Express the new wavelength in terms of the initial wavelength
Substituting \( v_{\text{total}} \) into the equation for \( \lambda_t \):
\[ \lambda_t = \frac{h}{m \sqrt{5v_0^2 + \left(\frac{qE_0}{m} t\right)^2}} \]
Now, we can express this in terms of \( \lambda_0 \):
\[ \lambda_t = \lambda_0 \cdot \frac{1}{\sqrt{5 + \frac{(qE_0 t)^2}{m^2 v_0^2}}} \]

### Final Result
Thus, the de-Broglie wavelength at time \( t \) is given by:
\[ \lambda_t = \lambda_0 \cdot \frac{1}{\sqrt{5 + \frac{(qE_0 t)^2}{m^2 v_0^2}}} \]