Given:

• Shortest wavelength of hydrogen atom in Lyman series is $\lambda$.
• We need the longest wavelength in Balmer series of He$^+$.

Shortest wavelength in Lyman series of H

For hydrogen-like species, $\frac{1}{\lambda}=RZ^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

In the Lyman series, $n_1=1$. The shortest wavelength corresponds to $n_2=\infty$.

So for hydrogen ($Z=1$): $\frac{1}{\lambda}=R\left(1-0\right)=R$ Hence, $\lambda=\frac{1}{R}$

Longest wavelength in Balmer series of He$^+$

For Balmer series, $n_1=2$. The longest wavelength corresponds to the smallest transition in that series, i.e. $n_2=3 \to n_1=2$.

For He$^+$, $Z=2$.

Therefore, $\frac{1}{\lambda'}=R(2)^2\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$ $\frac{1}{\lambda'}=4R\left(\frac{1}{4}-\frac{1}{9}\right)$ $\frac{1}{\lambda'}=4R\left(\frac{5}{36}\right)=\frac{5R}{9}$

Thus, $\lambda'=\frac{9}{5R}$

Use $\lambda=\frac{1}{R}$

Since $\lambda=\frac{1}{R}$, $\lambda'=\frac{9}{5}\lambda$

Match with options

$\boxed{\lambda'=\frac{9\lambda}{5}}$

So the correct option is C.