Question

An electron collides with a fixed hydrogen atom in its groundstate. Hydrogen atom gets excited and the colliding electron losesall its kinetic energy. Consequently the hydrogen atom may emit aphoton corresponding to the largest wavelength of the Balmer series. The K.E. of colliding electron will be

• A. 10.2 eV
• B. 1.9 eV
• C. 12.1 eV
• D. 13.6 eV

Answer 1

The correct option is C 12.1 eV

Solution:-

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. =$E_n=\frac{m_e{e}^4}{8n^2{E_o}^2{h}^2}$

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,

$E_n=\frac{-13.6}{n^2}eV$

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

$E_3=\frac{-13.6}{3^2}eV$

$E_3=-1.51eV⟶\left(1\right)$

Energy of $n=1$ Quantum state,

$E_1=\frac{-13.6}{1^2}eV$

$E_1=-13.6eV⟶\left(2\right)$

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

$K=E_3{-E}_1$

$K=-1.51-(-13.6)$

$K=12.1eV$