Physics - Modern Physics Question with Solution | TestHub

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 $n_i⟶n_f$ $$\begin{array}{rl}& \therefore \lambda \propto \frac{1}{Z^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)}\\ & \therefore \frac{{\lambda }_2}{{\lambda }_1}=\frac{Z_1^2{\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)}_1}{Z_2^2{\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)}_2}\\ & \end{array}$$ $${\lambda }_2=\frac{{\lambda }_1{Z}_1^2{\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)}_1}{Z_2^2{\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)}_2}$$ Substituting the values, we have $$=\frac{(6561\mathring{A})(1{)}^2\left(\frac{1}{2^2}-\frac{1}{3^2}\right)}{(2{)}^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right)}=1215\mathring{A}$$ $\therefore$ 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. $\mathrm{I}\to$ Lymen series II $\to$ Balmer series III $\to$ Paschen series IV $\to$ Brackett series $\mathrm{V}\to$ Pfund series