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JEE Mains Model Paper 8
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Question : 50
Total: 90
A photon collides with a stationary hydrogen atom in ground state and is completely absorbed (Perfectly inelastic collision). Energy of the colliding photon is 10.2 eV. After a micro second, anotherphoton collides with the same hydrogen atom perfectly in elastically with an energy of 15 eV. What will be observed by the detector ?(Neglect kinetic energy attained by hydrogen atom during the process)
Two photons of energy 1.2 eV
Two photons of energy 1.4 eV.
One photon of energy 10.2 eV and one electron of kinetic energy 1.4 eV.
One electron having kinetic energy nearly 11.6 eV.
Validate
Solution:
Total energy received by the atom will be 25.2 eV. 13.6 eV energy is needed to remove the electron fromthe attraction of the nucleus. Rest of the energy will be almost available in the form of KE of electron
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