Ask Questions, Get Answers

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

A stationary $He^+$ ion emitted a photon corresponding to the first line $(H_{\alpha})$ of the Lyman series .That photon liberated a photo electron from a stationary H atom in ground state.What is the velocity of photo electron?$R_H=109678cm^{-1}$

$\begin{array}{1 1}(a)\;2.03\times 10^8cm/s&(b)\;3.09\times 10^8cm/s\\(c)\;3.09\times 10^2cm/s&(d)\;1.03\times 10^8cm/s\end{array}$

Can you answer this question?

1 Answer

0 votes
Energy of photon liberated from $He^+$
$H_{\alpha}$ line of Lyman series =$hcR_HZ^2\big[\large\frac{1}{1^2}-\frac{1}{2^2}\big]$
$\Rightarrow 6.625\times 10^{-27}\times 3\times 3\times 10^{10}\times 109678\times 2^2\big[\large\frac{3}{4}\big]$
$\Rightarrow 6.54\times 10^{-11}erg$
$6.54\times 10^{-11}=E_1$ of H +$\large\frac{1}{2}$$mu^2$
$\Rightarrow 13.6\times 1.602\times 10^{-21}+\large\frac{1}{2}$$mu^2$
$\large\frac{1}{2}$$mu^2=6.54\times 10^{-11}-2.179\times 10^{-11}$
$\Rightarrow 4.361\times 10^{-11}erg$
$u^2=\large\frac{4.361\times 10^{-11}\times 2}{9.108\times 10^{-28}}$
$u=3.09\times 10^8cm/s$
Hence (b) 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