$\begin{array}{1 1}(a)\;n=2\;to\;n=3&(b)\;n=6\;to\;n=7\\(c)\;n=7\;to\;n=4&(d)\;n=4\;to\;n=7\end{array}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Electron transition from $n_2$ level is associated with wavelength $\lambda$ given by

$\large\frac{1}{\lambda}=$$R_H\big[\large\frac{1}{n_1^2}-\frac{1}{n_2^2}\big]$

$\lambda=2170nm=2170\times 10^{-9}m$

$R_H=1.09677\times 10^7m^{-1}$

$n_1=?n_2=7$

$\therefore \large\frac{10^9}{2170}=$$1.09677\times 10^7\big[\large\frac{1}{n_1^2}-\frac{1}{7^2}\big]$

$\big[\large\frac{1}{n_1^2}-\frac{1}{7^2}=$$0.0420$

$\large\frac{1}{n_1^2}$$=0.420+\large\frac{1}{49}$$=0.064$

$\therefore n_1=4$

Electron transition from $n=7$ to $n=4$ will produce infrared light of wavelength 2170nm

Hence (c) is the correct answer.

Ask Question

Tag:MathPhyChemBioOther

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...