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STD 12 - 12. atoms — NEET STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceNEETSTD 12 - 12. atoms4 Marks
Question
An electron collides with a fixed hydrogen atom in its ground state. Hydrogen atom gets excited and the colliding electron loses all its kinetic energy. Consequently the hydrogen atom may emit a photon corresponding to the largest wavelength of the Balmer series. The min. $K.E.$ of colliding electron will be.....$eV$
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Answer

Longest wavelength of the Balmer series for a hydrogen atom corresponds to a transition of an electron from $n=3$ Quantum state and $n=2$ quantum level. This means that when the hydrogen Atom was excited by the colliding electron, the electron went from $n=1$ to $n=3$ quantum state.

Energy of $n-n$ Quantum state for a hydrogen atom. $=E_{n}=\frac{m_{e} e^{4}}{8 n^{2} E_{o}^{2} h^{2}}$

Where,

$m_{e}=$ mass of electron.

$e=$ charge of an electron.

$n=$ quantum number(Orbit).

$E_{o}=$ absolute permittivity.

$h=$ Planck's Constant.

By evaluating the constants we get,

$E_{n}=\frac{-13.6}{n^{2}} e V$

$\therefore$ Energy of $n=3$ Quantum state,

$E_{3}=\frac{-13.6}{3^{2}} e V$

$E_{3}=-1.51 e V \longrightarrow(1)$

Energy of $n=1$ Quantum state,

$E_{1}=\frac{-13.6}{1^{2}} e V$

$E_{1}=-13.6 e V \longrightarrow(2)$

since this transition happens at the total Express of the kinetic energy of the colliding electron $K$ is:

$K=E_{3}-E_{1}$

$K=-1.51-(-13.6)$

$K=12.1 e V$

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