A) energy
B) momentum
C) velocity
D) angular momentum
Correct Answer: B
Solution :
de-Broglie wavelength is given by \[-{{Q}^{2}}(4\pi {{\varepsilon }_{0}}{{a}^{2}})\] where h is Plancks constant, p is momentum and \[{{Q}^{2}}/(2\pi {{\varepsilon }_{0}}{{a}^{2}})\], is wavelength. Since, electron and photon have the same wavelength the momentum of both will be same.
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