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An electron and a photon each have a wavelength of 1.00 nm. Find (a) their momentum , (b) the energy of the photon and (c) the kinetic energy of electron. |
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Answer» SOLUTION :Here `lambda=1.00nm =1XX10^(-9)` m. (a) Wavelength of electron and photon is same hence according to `p=(h)/(lambda)` their momentum will be equal `therefore` Equal momentum, `p=(h)/(lambda)=(6.63xx10^(-34))/(1xx10^(-9))` `therefore p=6.63xx10^(-25)kg ms^(-1)` (b)Energy of photon, `E=(hc)/(lambda)=(6.63xx10^(-34)xx3xx10^(8))/(1xx10^(-9))` `therefore E=19.89xx10^(-17)J` `therefore E=(19.89xx10^(-17))/(1.6xx10^(-19))EV` `therefore E=12.43xx10^(2)eV` `~~1.243.10^(3)eV` (C )KINETIC energy of electron, `K=(p^(2))/(2m)=((6.63xx10^(-25))^(2))/(2xx9.1xx10^(-31))=(43.9569xx10^(-5))/(18.2xx10^(-31))` `thereforeK=2.4152xx10^(-19)J` `therefore=(2.452xx10^(-19))/(1.6xx10^(-19))eV` `=1.509eV~~1.51 eV` |
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