A) A positron
B) A photon
C) An a-particle
D) A neutron
Correct Answer: C
Solution :
The de-Broglie equation is \[\lambda =\frac{h}{p}=\frac{h}{mv}\] Here, h and v are constant. So, \[\lambda \propto \frac{1}{m}\]. Since, the a-particle has the highest mass among the given entities, it has the smallest de-Broglie wavelength.You need to login to perform this action.
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