Question 33: Dual Nature of Matter and Radiation

Question

An electron of mass $m$ is moving in an electric field $\vec{E}=-2 E_{\mathrm{o}} \hat{i}\left(E_{\mathrm{o}}=\right.$ constant $\left.>0\right)$, with an initial velocity $\vec{V}=v_{\mathrm{o}} \hat{i} \left(v_{\mathrm{o}}=\right.$ constant $\left.>0\right)$. If $\lambda_{\mathrm{o}}=\frac{h}{4 m v_{\mathrm{o}}}$, its de Broglie wavelength at time $t$ is
$\_\_\_\_$ .
( $e=$ charge of electron)

Accepted Answer

For an electron, charge is $q=-e$.
Given electric field,
$$ \vec E=-2E_0 \hat i $$
Force on the electron is
$$ \vec F=q\vec E=(-e)(-2E_0 \hat i)=2eE_0 \hat i $$
So acceleration is
$$ \vec a=\frac{\vec F}{m}=\frac{2eE_0}{m}\hat i $$
Since initial velocity is along $\hat i$,
$$ \vec v(0)=v_0 \hat i $$
therefore velocity at time $t$ will be
$$ \vec v(t)=\left(v_0+\frac{2eE_0}{m}t\right)\hat i $$
Now de Broglie wavelength is
$$ \lambda=\frac{h}{p}=\frac{h}{mv} $$
So at time $t$,
$$ \lambda(t)=\frac{h}{m\left(v_0+\frac{2eE_0}{m}t\right)} $$
Given
$$ \lambda_0=\frac{h}{4mv_0} $$
So,
$$ \frac{h}{mv_0}=4\lambda_0 $$
Hence,
$$ \lambda(t)=\frac{\frac{h}{mv_0}}{1+\frac{2eE_0}{m}\frac{t}{v_0}} =\frac{4\lambda_0}{1+\frac{2eE_0}{m}\frac{t}{v_0}} $$
Therefore the correct answer is
$$ \boxed{\frac{4\lambda_0}{\left[1+\frac{2E_0e}{m}\frac{t}{v_0}\right]}} $$
So,
Option C
is correct.

Answer

$$ \frac{4 \lambda_{\mathrm{o}}}{\left[1-\frac{E_{\mathrm{o}} e}{2 m} \frac{t}{v_{\mathrm{o}}}\right]} $$

Answer

$$ \frac{4 \lambda_{\mathrm{o}}}{\left[1+\frac{E_{\mathrm{o}} e}{2 m} \frac{t}{v_{\mathrm{o}}}\right]} $$

Answer

$$ \frac{4 \lambda_{\mathrm{o}}}{\left[1+\frac{2 E_{\mathrm{o}} e}{m} \frac{t}{v_{\mathrm{o}}}\right]} $$

Answer

$$ \frac{4 \lambda_{\mathrm{o}}}{\left[1-\frac{2 E_{\mathrm{o}} e}{m} \frac{t}{v_{\mathrm{o}}}\right]} $$

Dual Nature of Matter and Radiation

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