JEE MAIN - Chemistry (2023 - 31st January Morning Shift - No. 11)
Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from $$\mathrm{n=4}$$ to $$\mathrm{n}=2$$ of $$\mathrm{He}^{+}$$ spectrum
$$\mathrm{n}=3$$ to $$\mathrm{n}=4$$
$$\mathrm{n}=2$$ to $$\mathrm{n}=1$$
$$\mathrm{n}=1$$ to $$\mathrm{n}=2$$
$$\mathrm{n}=1$$ to $$\mathrm{n}=3$$
Explanation
$\mathrm{He}^{+}$ion :
$$ \begin{aligned} & \frac{1}{\lambda(\mathrm{H})}=\mathrm{R}(1)^2\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\\\ & \frac{1}{\lambda\left(\mathrm{He}^{+}\right)}=\mathrm{R}(2)^2\left[\frac{1}{2^2}-\frac{1}{4^2}\right] \\\\ & \text { Given } \lambda(\mathrm{H})=\lambda\left(\mathrm{He}^{+}\right) \\\\ & \mathrm{R}(1)^2\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]=\mathrm{R}(4)\left[\frac{1}{2^2}-\frac{1}{4^2}\right] \\\\ & \frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}=\frac{1}{1^2}-\frac{1}{2^2} \end{aligned} $$
On comparing $\mathrm{n}_1=1 $ and $ \mathrm{n}_2=2$.
$$ \begin{aligned} & \frac{1}{\lambda(\mathrm{H})}=\mathrm{R}(1)^2\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\\\ & \frac{1}{\lambda\left(\mathrm{He}^{+}\right)}=\mathrm{R}(2)^2\left[\frac{1}{2^2}-\frac{1}{4^2}\right] \\\\ & \text { Given } \lambda(\mathrm{H})=\lambda\left(\mathrm{He}^{+}\right) \\\\ & \mathrm{R}(1)^2\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]=\mathrm{R}(4)\left[\frac{1}{2^2}-\frac{1}{4^2}\right] \\\\ & \frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}=\frac{1}{1^2}-\frac{1}{2^2} \end{aligned} $$
On comparing $\mathrm{n}_1=1 $ and $ \mathrm{n}_2=2$.
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