Nuclear Physics And Radioactivity Question 37
Question: A double charged lithium atom is equivalent to hydrogen whose atomic number is 3. The wavelength of required radiation for emitting electron from first to third Bohr orbit in $ L{{i}^{++}} $ will be (Ionisation energy of hydrogen atom is 13.6eV) [UPSEAT 1999]
Options:
A) 182.51 Å
B) 177.17 Å
C) 142.25 Å
D) 113.74 Å
Show Answer
Answer:
Correct Answer: D
Solution:
$ E_{n}=-13.6\frac{Z^{2}}{n^{2}}eV. $ Required energy for said transition $ \Delta E=E_{3}-E_{1}=13.6\ Z^{2}[ \frac{1}{1^{2}}-\frac{1}{3^{2}} ] $
$ \Rightarrow \Delta E=13.6\times 3^{2}[ \frac{8}{9} ]=108.8\ eV $
$ \Rightarrow \Delta E=108.8\times 1.6\times {{10}^{-19}}J $ Now $ \Delta E=\frac{hc}{\lambda }=108.8\times 1.6\times {{10}^{-19}} $
$ \Rightarrow \lambda =\frac{6.6\times {{10}^{-34}}\times 3\times 10^{8}}{108.8\times 1.6\times {{10}^{-19}}}=0.11374\times {{10}^{-7}}m $ $ =113.74{\AA} $
BETA
