In hydrogen atom ${\text{E}}_{\text{n}}-\genfrac{}{}{0.1ex}{}{Rhc}{n^2}$

where $m$ is the mass of the electron. Here, the electron has been replaced by a particle whose mass is double of an electron. Therefore, energy in $n^{\mathrm{th}}$ orbit for this hypothetical atom will be given by

$E_n=-\genfrac{}{}{0.1ex}{}{2Rhc}{n^2}$

The longest wavelength $\lambda$max (or minimum energy) photon will correspond to the transition of the particle from $n = 3$ to $n = 2$

$\frac{\mathit{\text{hc}}}{{\lambda }_{\text{max}}}={\mathit{\text{E}}}_3-{\mathit{\text{E}}}_2=2\mathit{\text{Rhc}}\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$

This gives, λ _max$=\genfrac{}{}{0.1ex}{}{18}{\mathrm{}5R}$