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 :
Wavelength,\[\lambda =\frac{h}{\sqrt{mE}}\] \[\Rightarrow \] \[E=\frac{{{h}^{2}}}{2m{{\lambda }^{2}}}\] \[\lambda \]is same for all, so\[E\propto \frac{1}{m}\].Hence, energy will be maximum for particle with lesser mass.
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