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Consider two hydrogen atoms H_(A)andH_(B) in ground state. Assume that hydrogen atom H_(A) is at rest and hydrogen atom H_(B) is moving with a speed and make head-on collide on the stationary hydrogen atom H_(A). After the strike, both of them move together. What is minimum value of the kinetic energy of the moving hydrogen atom H_(B), such that any one of the hydrogen atoms reaches one of the excitation state. |
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Answer» Solution :Collision between hydrogen `H_(A)` and hydrogen `H_(B)` atom will be inelastic if a part of kineticenergy is used to EXCITE atom If `u_(1)` and `u_(2)` are speed of `H_(A)` and `H_(B)` atom after collision then mu = m`u_(1)` + m`u_(2)`...(1) `(1)/(2)` m`U^(2) = (1)/(2)` m`u_(1)^(2) + (1)/(2)` m`u_(2)^(2)` + DELTA E ....(2) `u^(2) = u_(1)^(2) + (u -u_(1))^(2) + (2 Delta E)/(m)` `u_(1)^(2) - uu_(1) + (2 DeltaE)/(m) = 0` For `u_(1)` to be real `u^(2) - (4 Delta E)/(m) ge 0` (m`u^(2))/(2) ge 2 xx Delta E` `Delta E = 10.2 eV` THUS (1)/(2) m`u^(2)_("min") = 2 xx 10.2 eV` `((1)/(2)` m`u^(2))_("min") = 20.4 eV` The minimum K.E of the moving hydrogen atom `H_(B) is 20.4 eV` |
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