$\begin{array}{1 1}(a)\;10.2eV,2&(b)\;9.8eV,3\\(c)\;81.3eV,4&(d)\;5.3eV,1\end{array}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

Pls answer this question asap

0 votes

$E_1$ for H=-13.6eV

$E_2$ for $H=\large\frac{-13.6}{2^2}=\frac{13.6}{4}$$=-3.4eV$

$E_2-E_1=-3.4-(-13.6)=10.2eV$

Also for transition of H like atom

$\lambda=3\times 10^{-8}m$

$\large\frac{1}{\lambda}$$=R_H\times z^2\bigg[\large\frac{1}{1^2}-\frac{1}{2^2}\bigg]$

$R_H=109677cm^{-1}=109677\times 10^2m^{-1}$

$\large\frac{1}{3\times 10^{-8}}$$=109677\times 10^2\times z^2\big[\large\frac{3}{4}\big]$

$z^2=4$

$z=2$

Hence (a) is the correct answer.

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...