When an electron with energy 10.8 ,eV collide | vedclass

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(C) The ground state energy of a hydrogen atom is $-13.6 \,eV$. The energy required to excite an electron from the $n=1$ to $n=2$ state is $\Delta E =

Vedclass · Accepted Answer

The $H$ atom will absorb $10.2 \,eV$ of energy,and the electron will be emitted with $0.6 \,eV$ of energy.

Vedclass · Answer

(C) The ground state energy of a hydrogen atom is $-13.6 \,eV$. The energy required to excite an electron from the $n=1$ to $n=2$ state is $\Delta E = E_2 - E_1 = -3.4 \,eV - (-13.6 \,eV) = 10.2 \,eV$. Since the incident electron has $10.8 \,eV$ of energy,the $H$ atom absorbs $10.2 \,eV$ to undergo excitation,and the remaining energy $(10.8 \,eV - 10.2 \,eV = 0.6 \,eV)$ is retained by the electron as kinetic energy.

When an electron with energy $10.8 \,eV$ collides with a hydrogen $(H)$ atom,what will happen?

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