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Question

A hydrogen atom at rest is in ground state. It is struck by a He ion in first excited state. Assuming the collision to be head on and the mass of He+ to be four times that of hydrogen atom, find the least value of kinetic energy of incoming particle which can excite both the particles to second excited state.

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Solution

We know that, ϵn=13.6×z2n2

Ground state energy of H=13.6×1212=13.6eV

second excited state energy of H=13.6×1232

=13.69=1.52eV
Now,
First excited state of He+=13.6×2222

=13.6eV

second excited state of Net He+=13.6×2232eV=13.6×49eV=6.04eV

Now, Let KE be the minimum kinetic energy of He+ ion, So that both single electron species get excited, then

(Ground state energy of H)+(First excited state energy of He+)+KE=(2nd excited state energy of H)+(2nd excited state energy of He+)

(13.6)+(13.6)+KE=(1.52)+(6.04)eV

27.2+KE=7.56

KE=7.56+27.20

KE=+19.64eV

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