Nuclear Physics And Radioactivity Question 82
Question: Consider a hydrogen like atom whose energy in nth exicited state is given by $ E_{n}=-\frac{13.6Z^{2}}{n^{2}} $ when this excited atom makes a transition from excited state to ground state, most energetic photons have energy Emax = 52.224 eV and least energetic photons have energy Emin = 1.224 eV. The atomic number of atom is
Options:
A) 2
B) 5
C) 4
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Maximum energy is liberated for transition $ E_{n}\to 1 $ and minimum energy for $ E_{n}\to {E_{n-1}} $ Hence $ \frac{E_{1}}{n^{2}}-E_{1}=52.224,eV $ ??(i) and $ \frac{E_{1}}{n^{2}}-\frac{E_{1}}{{{(n-1)}^{2}}}=1.224,eV $ ?..(ii) Solving equations (i) and (ii) we get $ E_{1}=-54.4,eV $ and $ n=5 $ Now $ E_{1}=-\frac{13.6,Z^{2}}{1^{2}}=-54.4,eV. $ Hence $ Z=2 $
BETA
