AP EAMCET · PHYSICS · Dual Nature of Matter
An electron of mass ' \(m\) ' with initial velocity \(\vec{v}=v_0 \hat{i}\left(v_0\gt0\right)\) enters in an electric field \(\overrightarrow{\mathrm{E}}=-\mathrm{E}_0 \hat{i} \quad\left[\mathrm{E}_0\right.\) is constant \(\left.\gt0\right]\) at \(t=0\). If \(\lambda\) is its de-Brogli wavelength initially, then the de-Brogli wave length after time ' \(t\) ' is
- A \(\frac{\lambda}{1+\frac{e \mathrm{E}_o t}{m v_0}}\)
- B \(\frac{\lambda}{\left(1-\frac{e \mathrm{E}_o t}{m v_0}\right)^2}\)
- C \(\left(1-\frac{e \mathrm{E}_o t}{m v_0}\right) \lambda\)
- D \(\left(1+\frac{e \mathrm{E}_o t}{m v_0}\right)^2 \lambda\)
Answer & Solution
Correct Answer
(A) \(\frac{\lambda}{1+\frac{e \mathrm{E}_o t}{m v_0}}\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{v_1}=v_0 \hat{i}, \vec{E}=-E_0 \hat{i}, \lambda_1=\lambda\) \(\therefore \quad\) Acceleration of electron, \(\vec{a}=\frac{\vec{F}}{m}=\frac{-\mathrm{e} \vec{E}}{m}=\left(\frac{\mathrm{eE}_{\mathrm{o}}}{\mathrm{m}}\right) \hat{\mathrm{i}}\) \(\therefore \quad\)…
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