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QJEEM23Ch5f03: JEE Main 2023 Chemistry - Class 11 JEEM23Ch5f03 Solution: Which transition in the hydrogen spectrum would have the same…

Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from n=4\mathrm{n=4} to n=2\mathrm{n}=2 of He+\mathrm{He}^{+} spectrum

Multiple Choice Options

  • A n=3\mathrm{n}=3 to n=4\mathrm{n}=4
  • B n=2\mathrm{n}=2 to n=1\mathrm{n}=1
  • C n=1\mathrm{n}=1 to n=2\mathrm{n}=2
  • D n=1\mathrm{n}=1 to n=3\mathrm{n}=3

Step-by-Step Verified Solution

He+\mathrm{He}^{+} ion :

amp;1λ(H)=R(1)2[1n12−1n22]amp;1λ(He+)=R(2)2[122−142]amp; Given λ(H)=λ(He+)amp;R(1)2[1n12−1n22]=R(4)[122−142]amp;1n12−1n22=112−122\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 n1=1\mathrm{n}_1=1 and n2=2\mathrm{n}_2=2 .

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