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The minimum kinetic energy required for ionization of a hydrogen atom is `E_(1)` in case electron is collided with hydrogen atom , it is `E_(2)` if the the hydrogen ion is collided and `E_(1)` when helium ion collided . Then.A. `E_(1) = E_(2) = E_(3) `B. `E_(1) gt E_(2) gt E_(3) `C. `E_(1) lt E_(2) lt E_(3) `D. `E_(1) gt E_(3) gt E_(2) ` |
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Answer» Correct Answer - C Assuming that ionization occurs as a result of a completely inelastic collision, we can write `m v - 0 = (m + m_(H)) u` where `m` is that mass of incident partical , `m_(H)` the mass of hydrogen atom , `v_(0)` the initial velocity of incident particals , and `u` the final common velocity of the partical after collision. Prior to collision, the `KE` of the incident partical was `E_(0) = (m v_(0)^(2))/(2)` The total kinetic energy after collision `E = ((m + m_(H)) u^(2))/(2) = (m^(2)v_(0)^(2))/(2 (m + m_(H)))` The decrease kinetic energy must be eqaul to ionization energy. Therefore `E_(1) = E_(0) - E = ((m_(H))/(m + m_(H))) E_(0)` i.e., `(E_(1))/(E_(0)) = (1)/(1 + (m)/(m_(H))` i.e., the greater the mass `m`, the smaller the fraction of initial kinetic energy that be used for ionization. |
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