Nuclear Physics And Radioactivity Question 79
Question: A hydrogen like atom of atomic number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, a photon of energy 40.8 eV is emitted. The value of n will be [Based onIIT-JEE (Mains) 2000]
Options:
A) 1
B) 2
C) 3
D) 4
Show Answer
Answer:
Correct Answer: B
Solution:
Let ground state energy (in eV) be $ E_{1} $ Then from the given condition $ E_{2n}-E_{1}=204,eV $ or $ \frac{E_{1}}{4n^{2}}-E_{1}=204,eV $
Þ $ E_{1}( \frac{1}{4n^{2}}-1 )=204,eV $ ?..(i) and $ E_{2n}-E_{n}=40.8,eV $
Þ $ \frac{E_{1}}{4n^{2}}-\frac{E_{1}}{n^{2}}=E_{1}( -\frac{3}{4n^{2}} )=40.8,eV $ ?..(ii) From equation (i) and (ii), $ \frac{1-\frac{1}{4n^{2}}}{\frac{3}{4n^{2}}}=5 $
Þ $ n=2 $
BETA
