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The de Broglie wavelength of an electron and the wavelength of a photon are same. The ratio between the energy of the photon and the momentum of the electron isA. `h`B. `c`C. `1//h`D. `1//c` |
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Answer» Correct Answer - B `lambda_(e) = lambda_(ph), (E_(ph))/(m_(e)V) =(E_(ph))/(((h)/(lambda_(e)))) =((hc)/(lambda_(ph)))/((h)/(lambda_(e))) =C` |
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