Structure Of Atom Question 11
Question 11 - 31 January - Shift 1
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
(1) $n=2$ to $n=1$
(2) $n=1$ to $n=3$
(3) $n=1$ to $n=2$
(4) $n=3$ to $n=4$
Show Answer
Answer: (1)
Solution:
Formula: Wavelength of Emitted Photon
$He^{+}$ion :
$\frac{1}{\lambda(H)}=R(1)^{2}[\frac{1}{n_1^{2}}-\frac{1}{n_2^{2}}]$
$\frac{1}{\lambda(He^{+})}=R(2)^{2}[\frac{1}{2^{2}}-\frac{1}{4^{2}}]$
Given $\lambda(H)=\lambda(He^{+})$
$R(1)^{2}[\frac{1}{n_1^{2}}-\frac{1}{n_2^{2}}]=R(4)[\frac{1}{2^{2}}-\frac{1}{4^{2}}]$
$\frac{1}{n_1^{2}}-\frac{1}{n_2^{2}}=\frac{1}{1^{2}}-\frac{1}{2^{2}}$
On comparing $n_1=1 \& n_2=2$
Ans. 1
