MCQ
The shortest wavelength of $He^+$ ion in Balmer series is $x$, then longest wavelength in the paschen series of $Li^{2+}$ is
- A$\frac{{36x}}{5}$
- ✓$\frac{{16x}}{7}$
- C$\frac{{9x}}{5}$
- D$\frac{{5x}}{9}$
$\frac{1}{\lambda}=R Z^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$
For shortest wavelength $x$ of Paschen series $n_{2}=\infty$ and $n_{1}=3$
$\frac{1}{x}=R(3)^{2}\left[\frac{1}{3^{2}}-\frac{1}{\infty^{2}}\right]$
$\Rightarrow x=\frac{9}{9 R}=\frac{1}{R}$
For longest wavelength of Paschen series $n_{2}=4$ and $n_{1}=3$
$\frac{1}{\lambda}=R(3)^{2}\left[\frac{1}{3^{2}}-\frac{1}{4^{2}}\right]$
$\Rightarrow \frac{1}{\lambda}=R(3)^{2}\left[\frac{(16-9)}{16 \times 9}\right]$
$\Rightarrow \lambda=\frac{16}{7 R}=\frac{16 x}{7}$
Hence, the correct option is $B$
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$(1)$ $C{H_2} = C{H_2}$
$(2)$ ${(C{H_3})_2}C = C{H_2}$
$(3)$ $C{H_3}CH = CHC{H_3}$
$(a)\,B{r_2}(l) \to B{r_2}(g)$
$(b)\,{H_2}O(s) \to {H_2}O(g)$
$(c)\,{N_2}\,\left[ {1\,atm,\,{{100}\,^o}C} \right] \to {N_2}\,\left[ {1\,atm,\,{{150}\,^o}C} \right]$
$(d)\,{N_2}\,(g) + 3{H_2}(g) \to 2N{H_3}(g)$
$(e)\,CaC{O_3}(s) \to CaO(s) + C{O_2}(g)$