Question

In a region where electric field exists as \[ \vec{E} = -E_0 \hat{i} \] Initially at \(t=0\), velocity of particle of mass \(m\) is \(4v_0 \hat{i\). If de-Broglie wavelength \(\lambda_0 = \dfrac{h}{4mv_0}\), find wavelength \(\lambda\) at time \(t\).}

• A. \(\displaystyle \lambda=\frac{h\lambda_0}{h+qE_0\lambda_0 t}\)
• B. \(\displaystyle \lambda=\frac{h\lambda_0}{h-qE_0\lambda_0 t}\)
• C. \(\displaystyle \lambda=\frac{h\lambda_0}{h+2E_0q\lambda_0 t}\)
• D. \(\displaystyle \lambda=\frac{h\lambda_0}{h+\frac{qE_0\lambda_0 t}{2}}\)

Answer 1

The Correct Option is B

Concept: De-Broglie wavelength \[ \lambda = \frac{h}{mv} \] Step 1: Acceleration due to electric field \[ F = qE \] \[ a = \frac{qE_0}{m} \] Since field is in negative \(x\)-direction, velocity decreases. Step 2: Velocity at time \(t\) \[ v = 4v_0 - \frac{qE_0}{m}t \] Step 3: Substitute in de-Broglie relation \[ \lambda = \frac{h}{m(4v_0 - \frac{qE_0}{m}t)} \] \[ \lambda = \frac{h}{4mv_0 - qE_0 t} \] Step 4: Use given relation \[ \lambda_0 = \frac{h}{4mv_0} \] Thus \[ \lambda = \frac{h\lambda_0}{h - qE_0\lambda_0 t} \] Hence correct option is (2). \[ \boxed{\lambda=\frac{h\lambda_0}{h-qE_0\lambda_0 t}} \]