InterviewSolution
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InterviewSolution
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An electron having energy `20 e V` collides with a hydrogen atom in the ground state. As a result of the colllision , the atom is excite to a higher energy state and the electron is scattered with reduced velocity. The atom subsequentily returns to its ground state with emission of rediation of wavelength `1.216 xx 10^(-7) m`. Find the velocity of the scattered electron. |
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Answer» The energy lost by the electron in exciting the hydrogen atom equals the energy corresponding to `lambda = 1.216 xx 10^(7) m` `hv = (hc)/(lambda) = (6.63 xx 10^(-34) xx 3.0 xx 10^(8))/(1.216 xx 10^(-7))` Now , the initial energy of electron . `E = 32 xx 10^(-19) J` Hence, the kinetic energy of the scattered electron, `E = 32 xx 10^(-19) J - 16.36 xx 10^(-19) J = 15.65 xx 10^(-19) J` The velocity `v` of the scattered electron is given by `(1)/(2) mv^(2) = E` or `v = ((2 E)/(m))^(1//2) = ((2 xx 15.64 xx 10^(-19))/(9.11 xx 10^(-31))) ^(1//2)` `= 1.86 xx 10^(6) ms^(-1)` |
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