Question

An electron with a speed of $2.1\times {10}^6\mathrm{m}/\mathrm{s}$ collides with a hydrogen atom, exciting the atom to the highest possible energy level. The atom then undergoes a quantum jump with $\mathrm{\Delta n}=1.$ What is the wavelength of the photon emitted in the quantum jump?

Short answer

The wavelength emitted is 656 nm.

Step-by-step solution

Given information

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

Determine the photon released during the quantum jump's wavelength

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.

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

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

${\mathrm{E}}_{1\to \mathrm{n}}\le \mathrm{T}$

Implies that,

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

That is,

$1-\frac{1}{{\mathrm{n}}^2}\le \frac{12.54}{13.60}=0.922$,

At the end, $\mathrm{n}\le 3.58$

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

For the wavelength, this produces,

localid="1651153926131" $\mathrm{\lambda }=\frac{\mathrm{hc}}{{\mathrm{E}}_{\mathrm{\gamma }}}=\frac{\mathrm{hc}}{13.6\mathrm{eV}\left(\frac{1}{2^2}-\frac{1}{3^2}\right)}=\frac{\left(4.135\times {10}^{-15}\right)\mathrm{eVs}\cdot \left(3\times {10}^8\right)\mathrm{m}/\mathrm{s}}{13.6\mathrm{eV}\left(\frac{1}{2^2}-\frac{1}{3^2}\right)}$

At the end,

$\mathrm{\lambda }=656\mathrm{nm}$