Question:
A parallel beam of electrons travelling in x - direction falls on a slit of width d (see figure). If after passing the slit, an electron acquires momentum \(p_y\) in the y - direction then for a majority of electrons passing through the slit (h is Planck's constant):
A parallel beam of electrons travelling in x - direction falls on a slit of width d (see figure). If after passing the slit, an electron acquires momentum \(p_y\) in the y - direction then for a majority of electrons passing through the slit (h is Planck's constant):
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Most electrons land in the central diffraction maximum, whose edge satisfies d sin(theta) = lambda = h/p.
Updated On: Nov 11, 2026
- \(|p_y|\,d \lt h\)
- \(|p_y|\,d \gt h\)
- \(|p_y|\,d \simeq h\)
- \(|p_y|\,d \gg h\)
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The Correct Option is C
Solution and Explanation
Electrons behave as waves of wavelength \(\lambda=h/p\). For a single slit, the central maximum (where most electrons arrive) extends to the first minimum at \(d\sin\theta=\lambda\).
The transverse momentum is \(p_y=p\sin\theta\), so \(p_y d = p\,d\sin\theta = p\lambda = h\).
Hence for a majority of electrons \(|p_y|\,d\simeq h\).
Answer: (3) \(|p_y|\,d\simeq h\)
The transverse momentum is \(p_y=p\sin\theta\), so \(p_y d = p\,d\sin\theta = p\lambda = h\).
Hence for a majority of electrons \(|p_y|\,d\simeq h\).
Answer: (3) \(|p_y|\,d\simeq h\)
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