MCQ
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
- ✓$n =2$ to $n =1$
- B$n =1$ to $n =3$
- C$n =1$ to $n =2$
- D$n =3$ to $n =4$
✓
Answer
Correct option: A.
$n =2$ to $n =1$
a
$He ^{+}$ion :
$He ^{+}$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 $\lambda( H )=\lambda\left( He ^{+}\right)$
$R (1)^2\left[\frac{1}{ n _1^2}-\frac{1}{ n _2^2}\right]= R (4)\left[\frac{1}{2^2}-\frac{1}{4^2}\right]$
$\frac{1}{ n _1^2}-\frac{1}{ n _2^2}=\frac{1}{1^2}-\frac{1}{2^2}$
On comparing $n _1=1$ and $n _2=2$
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