JEE Mains · Chemistry · STD 11 - 2. structure of atom
Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from \(n =4\) to \(n =2\) of \(He ^{+}\)spectrum
- A \(n =2\) to \(n =1\)
- B \(n =1\) to \(n =3\)
- C \(n =1\) to \(n =2\)
- D \(n =3\) to \(n =4\)
Answer & Solution
Correct Answer
(A) \(n =2\) to \(n =1\)
Step-by-step Solution
Detailed explanation
\(He ^{+}\)ion : \(\frac{1}{\lambda( H )}= R (1)^2\left[\frac{1}{ n _1^2}-\frac{1}{ n _2^2}\right]\) \(\frac{1}{\lambda\left( He ^{+}\right)}= R (2)^2\left[\frac{1}{2^2}-\frac{1}{4^2}\right]\) Given \(\lambda( H )=\lambda\left( He ^{+}\right)\)…
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