$\begin{array}{1 1}(a)\;32.63\times 10^{-19}J&(b)\;47.68\times 10^{-19}J\\(c)\;47.68\times 10^{-12}J&(d)\;40.6\times 40.6\times 10^{-19}J\end{array}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Energy absorbed =$\large\frac{hc}{\lambda}$

$\Rightarrow \large\frac{6.625\times 10^{-27}\times 3.0\times 10^{10}}{360\times 10^{-8}}$

$\Rightarrow 5.52\times 10^{-11}erg$

$\Rightarrow 5.52\times 10^{-18}J$

$\therefore E_{absorbed}$=E used in attractive forces +kinetic energy of electron

$\Rightarrow 5.52\times 10^{-18}-7.52\times 10^{-19}J$

$\Rightarrow 47.68\times 10^{-19}J$

Hence (b) is the correct answer.

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...