logo

Ask Questions, Get Answers

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

Estimate the difference in energy between first and second Bohr's orbit for a H atom.At what minimum atomic no a transition from $n=2$ to $n=1$ energy level would result in the emission of $X$-rays with $\lambda=3\times 10^8m$ ?which hydrogen atom like species does this atomic no corresponds to ?

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

Pls answer this question asap
Can you answer this question?
 
 

1 Answer

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.
answered Jan 15, 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
...