logo

Ask Questions, Get Answers

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

If the total energy of an electron in a hydrogen like atom in an excited state is -3.4eV then the debroglie wavelength of the electron is

$\begin{array}{1 1}(a)\;6.6\times 10^{-10}&(b)\;3\times 10^{-10}\\(c)\;5\times 10^{-9}&(d)\;9.3\times 10^{-12}\end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
Total energy=$\large\frac{-e^2}{2r_n}=$$-3.4eV=\large\frac{E_1}{n^2}$
$\therefore n^2=\large\frac{-13.6}{-3.4}$$=4$
$n=2$
Velocity=$\large\frac{u_1}{2}=\frac{2.18\times 10^8}{2}$$cm sec^{-1}$
$\therefore \lambda=\large\frac{h}{mu}$
$\Rightarrow \large\frac{6.6\times 10^{-27}\times 2}{9.108\times 10^{-28}\times 2.18\times 10^8}$
$\Rightarrow 6.6\times 10^{-10}$
Hence (a) is the correct answer.
answered Jan 10, 2014 by sreemathi.v
 

Related questions

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