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Question

The de Broglie wavelength of an electron moving with a velocity 1.5×108ms1 is equal to that of a photon. The ratio of the kinetic energy of the electron to that of the energy of photon is:

A
2
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B
4
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C
12
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D
14
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Solution

The correct option is D 14
Kparticle=12mV2 also λ=hmV
Kparticle=12(hλv).v2=vh2λ(1)
Kphoton=hcλ
KparticleKphoton=v2c=1.5×1082×3×102=14

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