91Ó°ÊÓ

The given electron speed is$\mathrm{T}=\frac{1}{2}{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^2$

We can approach this topic non-relativistically because the electron has a significantly slower speed than light. The electron's total energy is equal to its kinetic energy.

$\mathrm{T}=\frac{1}{2}{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^2$

The collision absorbs some of this energy, which is then utilised to excite the hydrogen atom. Assume the hydrogen atom is stimulated to a level of energy n. It must absorb because it was in the ground state at the start.

${\mathrm{E}}_{1→\mathrm{n}}=13.6\mathrm{eV}\left(1−\frac{1}{{\mathrm{n}}^2}\right)$

For ${\mathrm{E}}_{{\mathrm{n}}_1}$, because T cannot be higher than n, we must discover the highest possible.

${\mathrm{E}}_{1→\mathrm{n}}≤\mathrm{T}$

Implies that,

$13.6\mathrm{eV}\left(1−\frac{1}{{\mathrm{n}}^2}\right)≤\frac{1}{2}{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^2=12.54\mathrm{eV},$

That is,

$1−\frac{1}{{\mathrm{n}}^2}≤\frac{12.54}{13.60}=0.922$,

At the end, $\mathrm{n}≤3.58$

The maximum possible n is n=3, which causes the atom to be excited to its third energy level. It is deexcited $3→2$i.e., the emitted photon's energy is ${\mathrm{E}}_{\mathrm{γ}}={\mathrm{E}}_{2→3}=13.6\mathrm{eV}\left(\frac{1}{2^2}−\frac{1}{3^2}\right).$

For the wavelength, this produces,

localid="1651153926131" $\mathrm{λ}=\frac{\mathrm{hc}}{{\mathrm{E}}_{\mathrm{γ}}}=\frac{\mathrm{hc}}{13.6\mathrm{eV}\left(\frac{1}{2^2}−\frac{1}{3^2}\right)}=\frac{\left(4.135×{10}^{−15}\right)\mathrm{eVs}â‹\dots \left(3×{10}^8\right)\mathrm{m}/\mathrm{s}}{13.6\mathrm{eV}\left(\frac{1}{2^2}−\frac{1}{3^2}\right)}$

At the end,

$\mathrm{λ}=656\mathrm{nm}$