Discussion Forum

The wavelength of the first spectral line in the Balmer series of a hydrogen atoms is 6561 Å The wavelength of the second  spectral line in the Balmer series of singly ionized helium atoms is 

Answer:

Explanation:

  For hydrogen or hydrogen type atoms

 $\frac{1}{\lambda}=RZ^{2}\left(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right)$

 In the transition from $n_{i} \rightarrow n_{f}$

 $\therefore$    $\lambda \propto \frac{1}{Z^{2}\left(\frac{1}{n^{2}_{f}}-\frac{1}{n_{i}^{2}}\right)}$

  $\therefore$   $\frac{\lambda_{2}}{\lambda_{1}}=\frac{Z_{1}^{2}\left(\frac{1}{n^{2}_{f}}-\frac{1}{n_{i}^{2}}\right)_{1}}{Z_{2}^{2}\left(\frac{1}{n^{2}_{f}}-\frac{1}{n_{i}^{2}}\right)_{2}}$

$\lambda_{2}=\frac{\lambda_{1}Z_{1}^{2}\left(\frac{1}{n^{2}_{f}}-\frac{1}{n_{i}^{2}}\right)_{1}}{Z_{2}^{2}\left(\frac{1}{n^{2}_{f}}-\frac{1}{n_{i}^{2}}\right)_{2}}$

 Substituting the values , we have 

$Å \frac{(6561 Å)(1)^{2}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)}{(2)^{2}\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)}=1215$Å

$\therefore$   correct option is (a)

Analysis of Question

(i) Question is simple.
(ii) In modern physics, mostly questions are asked on the emission of a photon by the transition of electrons from some higher energy state to some lower energy state

(iii) For hydrogen and hydrogen-like atoms which have onlY single

electron, we can use the formula

   $\frac{1}{\lambda}=RZ^{2}\left(\frac{1}{n_{f}^{2}}-\frac{1}{n^{2}_{i}}\right)$

 (iv)  Further, the student should also remember different series, as shown

in figure below