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What should be the velocity of the electron so that its momentum equals of4000Å wavelength photon. |
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Answer» Solution :de-Broglie WAVELENGTH of electron `lambda = (h)/(P) = (h)/(mv)` `V = (h)/(m lambda)` `= (6.6 XX 10^(-34))/(9.11 xx 10^(-31)xx 4000 xx 10^(-10)) = (6.6 xx 10^(-34))/(36.44 xx 10^(-38)) = 0.18112 xx 10^(4)` `v = 1811 ms^(-1)` |
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