Question

If the wavelength of the first line of the Lyman series for the hydrogen atom is 1216$\mathring{A}$, then the wavelength of the first line of the Balmer series of the hydrogen spectrum is:

• A. $1216\mathring{A}$
• B. $6563\mathring{A}$
• C. $912\mathring{A}$
• D. $3648\mathring{A}$

Answer 1

Answer: (B)

$6563\mathring{A}$

$\begin{array}{c}\Delta E=\frac{hc}{\lambda }\\ ⟹\lambda =\frac{hc}{\Delta E}\\ \text{Where}E=-13.6z^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})\\ ⟹\lambda =\frac{hc}{-13.6z^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})}\\ \text{For lyman first line z=1;}{n}_1=1;n_2=2;\lambda =1216\dot{A}\\ ⟹1216\dot{A}=\frac{hc}{-13.6\left(1{)}^2\right(\frac{1}{1^2}-\frac{1}{2^2})}\\ ⟹1216\dot{A}=\frac{hc}{-13.6\times \left(\frac{3}{4}\right)}-\left(i\right)\\ \text{Given for first lime of balmar z=1;}{n}_2=3;n_1=2;\\ ⟹{\lambda }_b=\frac{hc}{-13.6(\frac{1}{2^2}-\frac{1}{3^2})}\\ ⟹{\lambda }_b=\frac{hc}{-13.6\left(\frac{5}{36}\right)}-\left(ii\right)\\ \text{On dividing (ii) by (i) we will get}{\lambda }_b\approx 6563\dot{A}\end{array}$