The ion He⁺ has one electron and Z = 2. Its jumps from n = 5, n = 6 and n = 7 to n = 4 form the…

Step 1. The rule. $\dfrac{1}{\lambda} = 4R\left(\dfrac{1}{16} - \dfrac{1}{n_i^2}\right)$.

Step 2. $5 \to 4$. $\dfrac{1}{16} - \dfrac{1}{25} = 0.0225$. So $\dfrac{1}{\lambda} = 4 \times 1.097 \times 10^{7}\ \text{m}^{-1} \times 0.0225 = 9.87 \times 10^{5}\ \text{m}^{-1}$ and $\lambda = 1013\ \text{nm}$.

Step 3. $6 \to 4$. $\dfrac{1}{16} - \dfrac{1}{36} = 0.03472$, so $\dfrac{1}{\lambda} = 1.524 \times 10^{6}\ \text{m}^{-1}$ and $\lambda = 656\ \text{nm}$.

Step 4. $7 \to 4$. $\dfrac{1}{16} - \dfrac{1}{49} = 0.04209$, so $\dfrac{1}{\lambda} = 1.847 \times 10^{6}\ \text{m}^{-1}$ and $\lambda = 541\ \text{nm}$.

Step 5. The match. The line at $656\ \text{nm}$ comes from the levels $6 = 2 \times 3$ and $4 = 2 \times 2$. So it matches the jump of hydrogen from $n = 3$ to $n = 2$.