The De Broglie wavelength of a particle is equal to that of a photon, then the energy of photon will be
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a
equal to the kinetic energy of the particle
b
less than the kinetic energy of particle
c
equal to the total energy of the particle
d
more than the kinetic energy of the particle
answer is D.
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Detailed Solution
for particle λ=hmv for particle, energy Epa=12mv2 for photon λph=hmphc for photon , energy Eph=mphc2 As λph=λpa⇒mphc=mv Now EphEpa=mphc212mv2=mphcc12mvv=2cv>1⇒Eph>Epa