Question 77: Dual Nature of Matter and Radiation

Question

An electron of mass ' m ' with an initial velocity $\overrightarrow{\mathrm{v}}=\mathrm{v}_0 \hat{i}\left(\mathrm{v}_0>0\right)$ enters an electric field $\overrightarrow{\mathrm{E}}=-\mathrm{E}_{\mathrm{o}} \hat{\mathrm{k}}$. If the initial de Broglie wavelength is $\lambda_0$, the value after time t would be

Accepted Answer

Derivation of the de Broglie wavelength after time $$t$$
The de Broglie wavelength is given by
$$ \lambda = \frac{h}{p}, $$
where $$h$$ is Planck's constant and $$p$$ is the momentum of the particle.
Initial Conditions:
Initially, the electron has a velocity
$$    \vec{v} = v_0\,\hat{i},    $$
so its momentum is
$$    p_0 = m v_0.    $$
The initial de Broglie wavelength is therefore
$$    \lambda_0 = \frac{h}{m v_0}.    $$
Effect of the Electric Field:
The electron enters an electric field
$$    \vec{E} = -E_0\,\hat{k}.    $$
For an electron with charge $$-e$$, the force is given by
$$    \vec{F} = -e\,\vec{E} = -e\left(-E_0\,\hat{k}\right) = e E_0\,\hat{k}.    $$
This force causes an acceleration in the $$\hat{k}$$ direction:
$$    a_z = \frac{F_z}{m} = \frac{e E_0}{m}.    $$
Momentum Components After Time $$t$$:
In the $$\hat{i}$$ direction, there is no force, so the momentum remains constant:
$$      p_x = m v_0.      $$
In the $$\hat{k}$$ direction, starting from rest, the momentum becomes:
$$      p_z = m v_z = m \left(a_z t\right) = e E_0 t.      $$
Total Momentum of the Electron:
The total momentum is the vector sum of the components:
$$    p = \sqrt{p_x^2 + p_z^2} = \sqrt{(m v_0)^2 + (e E_0 t)^2}.    $$
New de Broglie Wavelength:
Substituting the expression for the total momentum into the de Broglie relation gives:
$$    \lambda = \frac{h}{\sqrt{m^2 v_0^2 + e^2 E_0^2 t^2}}.    $$
Expressing this in terms of the initial de Broglie wavelength $$\lambda_0 = \frac{h}{m v_0}$$:
$$    \lambda = \frac{\lambda_0}{\sqrt{1+\frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}.    $$
This final expression matches the option:
$$ \frac{\lambda_0}{\sqrt{1+\frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}. $$

Answer

$\frac{\lambda_o}{\sqrt{1-\frac{\mathrm{e}^2 \mathrm{E}_{\mathrm{o}}^2 \mathrm{t}^2}{\mathrm{~m}^2 \mathrm{v}_{\mathrm{o}}^2}}}$

Answer

$\lambda_0$

Answer

$\frac{\lambda_o}{\sqrt{1+\frac{\mathrm{e}^2 \mathrm{E}_{\mathrm{o}}^2 \mathrm{t}^2}{\mathrm{~m}^2 v_o^2}}}$

Answer

$\lambda_{\mathrm{o}} \sqrt{1+\frac{\mathrm{e}^2 \mathrm{E}_{\mathrm{o}}^2 \mathrm{t}^2}{\mathrm{~m}^2 \mathrm{v}_{\mathrm{o}}^2}}$

Dual Nature of Matter and Radiation

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