Ask Questions, Get Answers

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

The ionization energy of a H like Bohr's atom is 4 Rydberg.Calculate the wavelength radiated when electron jumps from the first excited state to ground state.

$\begin{array}{1 1}(a)\;203.89A^{\large\circ}&(b)\;303.89A^{\large\circ}\\(c)\;209A^{\large\circ}&(d)\;114A^{\large\circ}\end{array}$

Can you answer this question?

1 Answer

0 votes
Energy of I orbit of H like atom =$4R_H$
$\Rightarrow 4\times 2.18\times 10^{-18}$ J
$E_1$ for H=$-2.18\times 10^{-18}J$
$E_H$ like atom =$E_{1H}\times z^2$
$-4\times 2.18\times 10^{-18}=-2.18\times 10^{-18}\times z^2$
(i.e) Atomic no of H like atom is 2 or it is $He^+$ for de-excitation of e in $He^+$ from $n_2=2$ to $n_1=1$
$E_2-E_1=3R_h=3\times 2.18\times 10^{-18}J$
$\therefore E_2-E_1=\large\frac{hc}{\lambda}$
$\lambda =\large\frac{6.625\times 10^{-34}\times 3.0\times 10^8}{3\times 2.18\times 10^{-18}}$
$\Rightarrow 303.89\times 10^{-10}m$
$\Rightarrow 303.89A^{\large\circ}$
Hence (b) is the correct answer.
answered Jan 17, 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