A stationary He^+ ion emitted a photon corresponding to a first line of the Lyman series. The photon liberated a photoelectron from a stationary H atom in the ground state. What is the velocity of the photoelectron?

Sep 7, 2017

The first line Lyman series is between $n = 1$ and $n = 2$. We know that ${\text{He}}^{+}$ is isoelectronic with $\text{H}$ atom, and thus, we can utilize the Rydberg equation for hydrogen-like atoms:

$\Delta E = - {Z}^{2} \cdot {R}_{H} \left(\frac{1}{n} _ {f}^{2} - \frac{1}{n} _ {i}^{2}\right)$

where:

• $\Delta E$ is the energy of the relaxed electron and thus of the emitted photon.
• $Z$ is the atomic number.
• ${R}_{H}$ is the Hydrogen ionization energy in the appropriate units (say, $\text{13.61 eV}$).
• ${n}_{k}$ is the principal quantum number for the $k$th electron state.

The energy that the emitted photon has due to the electronic relaxation is given by:

$\Delta E = - \left({2}^{2}\right) \cdot \left(\text{13.61 eV}\right) \left(\frac{1}{2} ^ 2 - \frac{1}{1} ^ 2\right)$

$=$ $\text{40.83 eV}$

and would be pertaining to the $\underline{2 p \to 1 s}$ relaxation (due to the selection rules requiring the total orbital angular momentum to change as $\Delta L = \pm 1$).

The stationary hydrogen atom in its ground state has an ionization energy of $\text{13.61 eV}$, so its photoelectron would have a kinetic energy of:

$\text{40.83 eV" - "13.61 eV" = "27.22 eV}$

or $4.361 \times {10}^{- 18} \text{J}$, using the conversion factor $\left(1.602 \times {10}^{- 19} \text{J")/("1 eV}\right)$.
(Why did I convert to $\text{J}$?)

As all electrons have mass, they also have velocities pertaining to the usual kinetic energy equation:

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

And thus, the forward velocity is given by:

$\textcolor{b l u e}{{v}_{e}} = \sqrt{\frac{2 K}{m} _ e}$

= sqrt((2 cdot 4.361 xx 10^(-18) cancel"kg" cdot "m"^2"/s"^2)/(9.109 xx 10^(-31) cancel"kg")

$= 3.094 \times {10}^{6} \text{m/s}$

$=$ $\underline{\textcolor{b l u e}{3.094 \times {10}^{8} \text{cm/s}}}$