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Question

A proton of mass $m$ with an initial velocity $\overrightarrow{V} = V_0 \hat{i} \left( V_0 > 0 \right)$ enters an electric field $\overrightarrow{E} = -E_0 \hat{i}$ ($E_0 =$ constant $> 0$). If $\lambda_0$ is its de-Broglie wavelength initially, then its de-Broglie wavelength at time $t$ is:

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: Option A
## CONCEPT FRAMEWORK
 **Primary Concept**: De Broglie wavelength and motion under constant acceleration
 **Related Topics**: Classical mechanics, electromagnetism, quantum mechanics
 **Prerequisites**: Understanding of de Broglie wavelength ($\lambda = \frac{h}{p}$), Newton's second law, kinematics.
## SOLUTION PATHWAY
**Given Information**:
  $m$: mass of proton
  $\overrightarrow{V} = V_0 \hat{i}$: initial velocity
  $\overrightarrow{E} = -E_0 \hat{i}$: electric field
  $\lambda_0$: initial de Broglie wavelength
  $E_0 > 0$: constant
  $V_0 > 0$
  $t$: time
**Method & Steps**:
 1. Find the acceleration of the proton.
  Reasoning: The force on the proton is given by $\overrightarrow{F} = q\overrightarrow{E} = e(-E_0 \hat{i}) = -eE_0 \hat{i}$, where $e$ is the charge of the proton. Using Newton's second law, $\overrightarrow{F} = m\overrightarrow{a}$, we get $m\overrightarrow{a} = -eE_0 \hat{i}$, so $a = -\frac{eE_0}{m}$.
  Formula/Rule: $\overrightarrow{F} = m\overrightarrow{a}$ 
 2. Find the velocity of the proton at time $t$.
  Working: Since the acceleration is constant, we can use the equation $v = u + at$, where $u$ is the initial velocity and $v$ is the final velocity. In this case, $v = V_0 - \frac{eE_0}{m}t$.
  Key Point: The velocity is along the x-axis only.
 3. Find the de Broglie wavelength at time $t$.
  Reasoning: The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$, where $h$ is Planck's constant and $p$ is the momentum.  Substituting the expression for $v$ from step 2, we get
$\lambda = \frac{h}{m(V_0 - \frac{eE_0}{m}t)} = \frac{h}{mV_0(1 - \frac{eE_0 t}{mV_0})} = \frac{h}{mV_0} \frac{1}{(1 - \frac{eE_0 t}{mV_0})} = \frac{\lambda_0}{(1 - \frac{eE_0 t}{mV_0})}$ since $\lambda_0 = \frac{h}{mV_0}$ 
  Result: $\lambda = \frac{\lambda_0}{(1 - \frac{eE_0 t}{mV_0})}$ 
  Verification: At $t=0$, $\lambda = \lambda_0$, which is consistent with the initial condition.
 ## OPTION ANALYSIS
A] ✓: Correctly derives the de Broglie wavelength using the velocity at time t.
B] :x: Incorrectly uses a linear approximation instead of the correct reciprocal relationship.
C] :x: Incorrectly uses an exponential relationship, which is not appropriate for constant acceleration.
D] :x: Incorrectly uses an exponential relationship with the wrong sign in the exponent.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Ignoring the vector nature of the quantities and using magnitudes directly.: Always consider the direction of the velocity and acceleration.
  2.Incorrectly applying kinematic equations.: Remember to use the appropriate kinematic equation for constant acceleration.
 **Quick Tips**:
  1.Remember the formula for de Broglie wavelength $\lambda = \frac{h}{p}$
  2.Always start by analyzing the forces and acceleration.

P.NARASIMHA · Answer

The correct Option is "$\frac{\lambda_0}{1 - \frac{eE_0 t}{mv_0}}$"

P.NARASIMHA · Answer

$\lambda_0(1 - \frac{eE_0 t}{mv_0})$

P.NARASIMHA · Answer

$\lambda_0 \exp(-\frac{eE_0 t}{mv_0})$

P.NARASIMHA · Answer

$\lambda_0 \exp(\frac{eE_0 t}{mv_0})$

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