1. Identify the Ion ($He^+$)

The problem states the spectral line energies range from a minimum of 1.224 eV. In a hydrogen-like atom (atomic number $Z$), the energy levels are given by $E_n = -13.6 \frac{Z^2}{n^2}$ eV.

The energy difference between levels $n_2$ and $n_1$ is ($n_2$ – $n_1$ should be 1):

$$ \Delta E = 13.6 Z^2 \left( \frac{1}{n_1^2} – \frac{1}{n_2^2} \right) $$

Checking for Helium ($He^+$, $Z=2$):
The transition $n=5 \to n=4$ (Brackett-$\alpha$ for He+):

$$ \Delta E = 13.6(2^2) \left( \frac{1}{16} – \frac{1}{25} \right) = 54.4 \left( \frac{9}{400} \right) = 1.224 \text{ eV} $$

This matches the minimum energy exactly. Therefore, the ion is $He^+$ ($Z=2$).

2. Calculate Photon Energy (2nd Balmer Line)

The Balmer series corresponds to transitions ending at $n_f = 2$.
1st line: $3 \to 2$
2nd line: $n_i = 4 \to n_f = 2$.

$$ E_{ph} = 13.6 Z^2 \left( \frac{1}{2^2} – \frac{1}{4^2} \right) = 54.4 \left( \frac{1}{4} – \frac{1}{16} \right) $$

$$ E_{ph} = 54.4 \left( \frac{3}{16} \right) = 10.2 \text{ eV} $$

3. Photoelectric Effect

The photon is incident on a metal with work function $\Phi = 2.20 \text{ eV}$. The maximum kinetic energy $K_{max}$ of the emitted photoelectrons is:

$$ K_{max} = E_{ph} – \Phi = 10.2 – 2.20 = 8.0 \text{ eV} $$

Convert to Joules:

$$ K_{max} = 8.0 \times 1.6 \times 10^{-19} = 12.8 \times 10^{-19} \text{ J} $$

4. Velocity Selector

The electrons pass undeviated through crossed electric ($E$) and magnetic ($B$) fields. This requires the electric force to balance the magnetic force: $qE = qvB \Rightarrow v = E/B$.

First, find the velocity $v$ from the kinetic energy:

$$ v = \sqrt{\frac{2K_{max}}{m}} = \sqrt{\frac{2 \times 12.8 \times 10^{-19}}{9.1 \times 10^{-31}}} = \sqrt{\frac{25.6 \times 10^{12}}{9.1}} $$

$$ v \approx \sqrt{2.813 \times 10^{12}} \approx 1.677 \times 10^6 \text{ m/s} $$

5. Calculate Electric Field Intensity

Given $B = 1.0 \text{ mT} = 1.0 \times 10^{-3} \text{ T}$.

$$ E_{field} = v B = (1.677 \times 10^6) \times (1.0 \times 10^{-3}) $$

$$ E_{field} = 1677 \text{ V/m} \approx 1.68 \text{ kV/m} $$