WBJEE · Physics · Dual Nature of Matter
A proton and an electron initially at rest are accelerated by the same potential difference. Assuming that a proton is 2000 times heavier than an electron, what will be the relation between the de Broglie wavelength of the proton \(\left(\lambda_{p}\right)\) and that of electron \(\left(\lambda_{c}\right) ?\)
- A \(\lambda_{p}=2000 \lambda_{e}\)
- B \(\lambda_{p}=\frac{\lambda_{e}}{2000}\)
- C \(\lambda_{p}=20 \sqrt{5} \lambda_{0}\)
- D \(\lambda_{p}=\frac{\lambda_{0}}{20 \sqrt{5}}\)
Answer & Solution
Correct Answer
(D) \(\lambda_{p}=\frac{\lambda_{0}}{20 \sqrt{5}}\)
Step-by-step Solution
Detailed explanation
As we know that de-Broglie wavelength is given as \(\lambda=\frac{h}{p}=\frac{h}{m v}\) where, \(h=\) Planck constant \[ \begin{array}{l} p=\text { momentum of particle } \\ v=\text { velocity of particle } \end{array} \] and \(m=\) mass of the particle. Eq.(i) can be written…
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