Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

The energy E for an electron in H atom is $\large\frac{21.7\times 10^{-12}}{h^2}$erg.Calculate the energy required to remove electron completely from $n=2$ orbit.Also calculate the longest wavelength of light that can be used to cause this transition.

$\begin{array}{1 1}(a)\;4367.4A^{\large\circ}&(b)\;3263A^{\large\circ}\\(c)\;3663.6A^{\large\circ}&(d)\;2348.1A^{\large\circ}\end{array}$

Can you answer this question?

1 Answer

0 votes
$E_n=-\large\frac{21.7\times 10^{-12}}{n^2}$erg
$\therefore E_2=-\large\frac{21.7\times 10^{-12}}{4}$$=5.425\times 10^{-12}erg$
For removal of electron $E_2=\large\frac{hc}{\lambda}$
$E_2$ should be given to remove electron (i.e) +ve
$\lambda=\large\frac{6.625\times 10^{-27}\times 3.0\times 10^{10}}{5.425\times 10^{-12}}$
$\Rightarrow 3663.6\times 10^{-8}cm$
$\Rightarrow 3663.6A^{\large\circ}$
Hence (c) is the correct answer.
answered Jan 16, 2014 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App