A photon collides with a stationary hydrogen atom in ground state inolestically . Energy of the order of inicro second another photon collicles with same hydrogen atom indastisically with an energy of `15 eV` what will be observad by the detanctor?
A
One photon of energy `10.2eV ` and an electronof energy `1.4 eV`
B
`2` photon of energy of `1:4 eV`
C
`2` photon of energy of `10.2 eV`
D
One photon of energy `10.2eV ` and another photon of `1.4 eV`
Text Solution
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The correct Answer is:
A
Initially a photon of energy `10.2eV` colllides ineltastically with a hydrogen atom in ground state for hydrogen atom, ` E_(1) = - 13.6 eV, E_(2) = -(13.6)/(4) eV = - 4eV` `:. E_(2) - E_(1) = 10.2 eV` The electron of hydrogen atom will jump to recound arbit after absurbing the photo of energy `10.2 eV` . The electronic jumps back to its original state its less then microseco and release a photon of energy `10.2 ev` . Another photon of energy `15 eV` strikes the hydrogen atom inelastically . This energy is ionisation energy is `13.6 eV` . The remain energy of `14eV in left with electron is its kinetic energy
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A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15eV. What wil be observed by the detector? (a) 2 photons of energy 10.2 eV (b) 2 photons of energy 1.4 eV (c ) One photon of energy 10.2 eV and an electron of energy 1.4 eV (d) One photon of energy 10.2 eV and another photon of energy 1.4 eV
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