The wavelength of the first spectral line in the Balmer series of hydrogen atom is $6561 Å$. The wavelength…
- $1215 Å$
- $1640 Å$
- $2430 Å$
- $4687 Å$
Solution
For hydrogen or hydrogen type atoms $ \frac{1}{\lambda}=R Z^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right) $ In the transition from $n_i \longrightarrow n_f$ $ \begin{aligned} & \therefore \quad \lambda \propto \frac{1}{Z^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)} \\ & \therefore \quad \frac{\lambda_2}{\lambda_1}=\frac{Z_1^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)_1}{Z_2^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)_2} \\ & \end{aligned} $ $ \lambda_2=\frac{\lambda_1 Z_1^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)_1}{Z_2^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)_2} $ Substituting the values, we have $ =\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 photon by the transition of electron 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}=R Z^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right) $ (iv) Further, the student should also remember different series, as shown in figure below. $\mathrm{I} \rightarrow$ Lymen series II $\rightarrow$ Balmer series III $\rightarrow$ Paschen series IV $\rightarrow$ Brackett series $\mathrm{V} \rightarrow$ Pfund series
Asked in: JEE Advanced 2011 (Paper 1)