Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
0 votes

A gas of identical hydrogen like atoms has some atoms in lowest (ground) energy level A and some atoms in a particular upper (excited) energy , level B and there are no atoms in any other energy level . The atoms of gas make transition to higher energy level by absorbing monochromatic light photon having energy 2.7eV . Subsequently the atoms emit radiation of only six different photon energies . Some of emitted photons have energy 2.7eV some have energy more and some have less than 2.7eV . Then the principal quantum no .of initially excited level B is


Can you answer this question?

1 Answer

0 votes
Answer : (b) 2
Explanation :
Level C must correspond to n=4 since total number of possible transitions from level 4 to lower levels is
The level B corresponds either to n=2 or n=3
The energies of all photons involved in transitions is (in eV)
$\varepsilon _{4.2}=13.6Z^2\;(\large\frac{1}{2^2}-\large\frac{1}{4^2})=2.55 Z^2$
$\varepsilon_{4,3}=0.66 Z^2$
$\varepsilon_{4,1}=12.75 Z^2$
$\varepsilon_{3,2}=1.89 Z^2$
$\varepsilon_{3,1}=12.1 Z^2$
$\varepsilon_{2,1}=10.2 Z^2$
The only possible choices are
a) Z=1 and $\;\varepsilon_{4,2}=2.55 eV \approx 2.7eV$
b) Z=1 and $\;\varepsilon_{4,3}=2.64 eV \approx 2.7eV$
The choice Z=2 however is inconsistent with the fact that some photons have energy less than 2.7eV
Thus , Z=1 and quantum number of level B is n2 .
answered Feb 25, 2014 by yamini.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