A) Is the photon
B) Is the electron
C) Is the uranium nucleus
D) Depends upon the wavelength and the properties of the particle.
Correct Answer: A
Solution :
\[\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2mE}}:\] \[\therefore \ E=\frac{{{h}^{2}}}{2m{{\lambda }^{2}}}\] \[\lambda \] is same for all, so \[E\propto \frac{1}{m}\]. Hence energy will be maximum for particle with lesser mass.
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