Q.35. An electron of mass m is moving in an electric field ( = constant 0), with an initial veloc

Q: What is the correct answer for this exam Physics Q35?

The correct answer is C. Sol. $\vec{V} = \vec{U} + \vec{a}t$. $\vec{F} = q\vec{E} = (-e)(-2E_0 \hat{i}) = 2eE_0 \hat{i}$. $\vec{a} = \frac{\vec{F}}{m} = \frac{2eE_0}{m} \hat{i}$. $\vec{V} = v_0 \hat{i} + \frac{2eE_0}{m} t \hat{i} = (v_0 + \frac{2eE_0}{m} t) \hat{i}$. The de Broglie wavelength $\lambda = \frac{h}{mV}$. So, $\lambda = \frac{h}{m(v_0 + \frac{2eE_0}{m} t)} = \frac{h}{mv_0 (1 + \frac{2eE_0}{mv_0} t)}$. Given $

Q: Q.35. An electron of mass m is moving in an electric field ( = constant > 0), with an initial velocity ( = constant > 0). If its de Broglie wavelength at time t is

The correct answer is C. Sol. $\vec{V} = \vec{U} + \vec{a}t$. $\vec{F} = q\vec{E} = (-e)(-2E_0 \hat{i}) = 2eE_0 \hat{i}$. $\vec{a} = \frac{\vec{F}}{m} = \frac{2eE_0}{m} \hat{i}$. $\vec{V} = v_0 \hat{i} + \frac{2eE_0}{m} t \hat{i} = (v_0 + \frac{2eE_0}{m} t) \hat{i}$. The de Broglie wavelength $\lambda = \frac{h}{mV}$. So, $\lambda = \frac{h}{m(v_0 + \frac{2eE_0}{m} t)} = \frac{h}{mv_0 (1 + \frac{2eE_0}{mv_0} t)}$. Given $

Q: What are the options for this this exam Physics question?

A. $\frac{4\lambda_0}{1 - \frac{2Ee t}{2m v_0}}$ · B. $\frac{4\lambda_0}{1 + \frac{Ee t}{2m v_0}}$ · C. $\frac{4\lambda_0}{1 + \frac{2Ee t}{m v_0}}$ · D. $\frac{4\lambda_0}{1 - \frac{2Ee t}{m v_0}}$

Question

Q.35. An electron of mass m is moving in an electric field  ( = constant > 0), with an initial velocity  ( = constant > 0). If  its de Broglie wavelength at time t is

FirstInTest · Accepted Answer

Sol. $\vec{V} = \vec{U} + \vec{a}t$. $\vec{F} = q\vec{E} = (-e)(-2E_0 \hat{i}) = 2eE_0 \hat{i}$. $\vec{a} = \frac{\vec{F}}{m} = \frac{2eE_0}{m} \hat{i}$. $\vec{V} = v_0 \hat{i} + \frac{2eE_0}{m} t \hat{i} = (v_0 + \frac{2eE_0}{m} t) \hat{i}$. The de Broglie wavelength $\lambda = \frac{h}{mV}$. So, $\lambda = \frac{h}{m(v_0 + \frac{2eE_0}{m} t)} = \frac{h}{mv_0 (1 + \frac{2eE_0}{mv_0} t)}$. Given $\lambda_0 = \frac{h}{4mv_0}$, so $h = 4mv_0 \lambda_0$. Substitute $h$: $\lambda = \frac{4mv_0 \lambda_0}{mv_0 (1 + \frac{2eE_0}{mv_0} t)} = \frac{4\lambda_0}{1 + \frac{2eE_0}{mv_0} t}$.

Answer

$\frac{4\lambda_0}{1 - \frac{2Ee t}{2m v_0}}$

Answer

$\frac{4\lambda_0}{1 + \frac{Ee t}{2m v_0}}$

Answer

$\frac{4\lambda_0}{1 + \frac{2Ee t}{m v_0}}$

Answer

$\frac{4\lambda_0}{1 - \frac{2Ee t}{m v_0}}$

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