A) proton
B) \[\alpha \]-particle
C) neutron
D) \[\beta \]-particle
Correct Answer: D
Solution :
de-Broglie wavelength is given by |
\[\lambda =\frac{h}{mv}\] |
For same velocity, |
\[\lambda \,\propto \frac{1}{m}\] |
Out of the given particles, the mass of \[\beta \]-particle which is a fast moving electron, is minimum. Thus, de-Broglie wavelength is maximum for \[\beta \]-particle. |
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