Question

The wavelength of first member of Balmer series of a hydrogen atom is nearly (The value of Rydberg constant $R=1.08\times {10}^7{m}^{-1}$

• A. $4400\mathring{A}$
• B. $5500\mathring{A}$
• C. $6600\mathring{A}$
• D. $7700\mathring{A}$

Answer 1

The correct option is C $6600\mathring{A}$

$\begin{array}{c}\text{we know, wavelength emitted is}\\ \frac{1}{\lambda }=R{z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})\end{array}$

$\begin{array}{c}\text{for, first member of balmer series}\\ \begin{array}{cc}{n}_1& =2\text{to}{n}_2=3\\ & z=1\left(\text{hydrogen}\right)\\ R& =1.08\times {10}^7{m}^{-1}\end{array}\end{array}$

$\begin{array}{cc}\frac{1}{\lambda }& =1.08\times {10}^7\times (1{)}^2(\frac{1}{4}-\frac{1}{9})\\ & =\frac{1.08\times {10}^7\times 5}{36}\\ \lambda & =\frac{36}{5\times 1.08\times {10}^{+7}}\\ \lambda & =6.66\times {10}^{-7}m\\ \lambda & =6600A\because 1\dot{A}={10}^{-10}m\end{array}$