An electron (mass m) with initial velocity `bar(v) = -v_(0)hati + 3v_(0)hatk` is in an electric field `hatE = = 2E_(0)hatj`. 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))/(sqrt(1 + (2e^(2)E_(0)^(2)t^(2))/(5m^(2)v_(0)^(2)))`
B. `(lamda_(0))/(sqrt(2 + (e^(2)E^(2)t^(2))/(m^(2)v_(0)^(2)))`
C. `(lamda_(0))/(sqrt(1 + (e^(2)E^(2)t^(2))/(2m^(2)v_(0)^(2)))`
D. `(lamda_(0))/(sqrt(1 + (e^(2)E_(0)^(2)t^(2))/(5m^(2)v_(0)^(2)))`

Explanation

To solve the problem, we need to find the de Broglie wavelength of an electron at a given time \( t \) after it has been subjected to an electric field. Here’s a step-by-step solution:

### Step 1: Determine the initial velocity of the electron
The initial velocity of the electron is given as:
\[ \vec{v}_0 = -v_0 \hat{i} + 3v_0 \hat{k} \]

### Step 2: Calculate the magnitude of the initial velocity
The magnitude of the initial velocity \( v_0 \) can be calculated using the Pythagorean theorem:
\[ |\vec{v}_0| = \sqrt{(-v_0)^2 + (3v_0)^2} = \sqrt{v_0^2 + 9v_0^2} = \sqrt{10v_0^2} = v_0 \sqrt{10} \]

### Step 3: Calculate the initial de Broglie wavelength
The de Broglie wavelength \( \lambda_0 \) is given by the formula:
\[ \lambda_0 = \frac{h}{mv} \]
Substituting the initial velocity:
\[ \lambda_0 = \frac{h}{m \cdot v_0 \sqrt{10}} = \frac{h}{m \sqrt{10} v_0} \]

### Step 4: Determine the force on the electron due to the electric field
The force \( \vec{F} \) acting on the electron in the electric field \( \vec{E} = 2E_0 \hat{j} \) is given by:
\[ \vec{F} = q \vec{E} \]
where \( q = -e \) (the charge of the electron). Therefore,
\[ \vec{F} = -e (2E_0 \hat{j}) = -2eE_0 \hat{j} \]

### Step 5: Calculate the acceleration of the electron
Using Newton's second law, the acceleration \( \vec{a} \) is:
\[ \vec{a} = \frac{\vec{F}}{m} = \frac{-2eE_0 \hat{j}}{m} \]

### Step 6: Find the velocity of the electron at time \( t \)
The velocity \( \vec{v}(t) \) at time \( t \) can be calculated using the equation:
\[ \vec{v}(t) = \vec{v}_0 + \vec{a}t \]
Substituting the values:
\[ \vec{v}(t) = (-v_0 \hat{i} + 3v_0 \hat{k}) + \left(-\frac{2eE_0}{m} \hat{j}\right)t \]
This gives:
\[ \vec{v}(t) = -v_0 \hat{i} + 3v_0 \hat{k} - \frac{2eE_0 t}{m} \hat{j} \]

### Step 7: Calculate the magnitude of the velocity at time \( t \)
The magnitude of the velocity is:
\[ |\vec{v}(t)| = \sqrt{(-v_0)^2 + \left(-\frac{2eE_0 t}{m}\right)^2 + (3v_0)^2} \]
This simplifies to:
\[ |\vec{v}(t)| = \sqrt{v_0^2 + \left(\frac{2eE_0 t}{m}\right)^2 + 9v_0^2} = \sqrt{10v_0^2 + \left(\frac{2eE_0 t}{m}\right)^2} \]

### Step 8: Calculate the de Broglie wavelength at time \( t \)
The de Broglie wavelength at time \( t \) is given by:
\[ \lambda(t) = \frac{h}{m |\vec{v}(t)|} \]
Substituting the expression for \( |\vec{v}(t)| \):
\[ \lambda(t) = \frac{h}{m \sqrt{10v_0^2 + \left(\frac{2eE_0 t}{m}\right)^2}} \]

### Step 9: Express \( \lambda(t) \) in terms of \( \lambda_0 \)
Using \( \lambda_0 = \frac{h}{m \sqrt{10} v_0} \):
\[ \lambda(t) = \frac{\lambda_0 \sqrt{10}}{\sqrt{10 + \frac{4e^2E_0^2 t^2}{m^2 v_0^2}}} \]
This can be simplified to:
\[ \lambda(t) = \lambda_0 \frac{1}{\sqrt{1 + \frac{4e^2E_0^2 t^2}{10v_0^2 m^2}}} \]

### Final Expression
Thus, the final expression for the de Broglie wavelength at time \( t \) is:
\[ \lambda(t) = \lambda_0 \frac{1}{\sqrt{1 + \frac{4e^2E_0^2 t^2}{10v_0^2 m^2}}} \]