So $\quad \lambda=\frac{ hc }{\Delta E } \quad$ for $\lambda$ minimum i.e.
shortest; $\Delta E=$ maximum
for Lyman series $n=1 \&$ for $\Delta E_{\text {max }}$
Transition must be form $n=\infty$ to $n=1$
So $\quad \frac{1}{\lambda}= R _{ H } Z^{2}\left(\frac{1}{ n _{1}^{2}}-\frac{1}{ n _{2}^{2}}\right)$3
$\frac{1}{\lambda}= R _{ H } Z^{2}(1-0)$
$\frac{1}{\lambda}= R \times(1)^{2} \Rightarrow \lambda_{1}=\frac{1}{ R }$
For longest wavelength $\Delta E =$ minimum for Balmer series $n =3$ to $n =2$ will have $\Delta E$ minimum
for $He ^{+} Z =2$
So $\frac{1}{\lambda_{2}}= R _{ H } \times Z^{2}\left(\frac{1}{ n _{1}^{2}}-\frac{1}{ n _{2}^{2}}\right)$
$\frac{1}{\lambda_{2}}= R _{ H } \times 4\left(\frac{1}{4}-\frac{1}{9}\right)$
$\frac{1}{\lambda_{2}}= R _{ H } \times \frac{5}{9}$
$\lambda_{2}=\lambda_{1} \times \frac{9}{5}$
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