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An electron and a photon possess the same de Broglie wavelength. If `E_e` and `E_ph` are, respectively, the energies of electron and photon while v and c are their respective velocities, then `(E_e)/(E_(ph))` is equal toA. `(v)/(c )`B. `(v)/(2c)`C. `(v)/(3c)`D. `(v)/(4c)` |
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Answer» Correct Answer - B `lamda=(h)/(sqrt(2mE_e))=(hc)/(E_(ph))` or `2mE_e=(E_(ph)^(2))/(c^(2))` but `E_(e)=(1)/(2)mv^(2)` or `m=(2E_(e))/(v^(2))` `2[(2E)/(v^(2))]E_(e)=(E_(ph)^(2))/(c^(2))` or `(4E_e^2)/(v^2)=(E_(ph)^2)/(c^2)` or `(E_e^2)/(E_(ph)^(2))=(v^2)/(4c^2)` or `(E_(e))/(E_(ph))=(v)/(2c)` |
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