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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
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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

$(a)\;1\qquad(b)\;2\qquad(c)\;3\qquad(d)\;4$

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Answer : (b) 2
Explanation :
Level C must correspond to n=4 since total number of possible transitions from level 4 to lower levels is
1+2+3=6
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
 

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