Neet Solved Paper 2015 Question 42
Question: Consider $ 3^{rd} $ orbit of $ H{e^{+}} $ (Helium), using non-relativistic approach, the speed of electron in this orbit will be (given $ K=9\times 10^{9} $ constant, Z = 2 and h (Planck’s constant) = $ 6.6\times {10^{-34}}J-s $ )
Options:
A) $ 2.92\times 10^{6}m/s $
B) $ 1.46\times 10^{6}m/s $
C) $ 0.73\times 10^{6}m/s $
D) $ 3.0\times 10^{8}m/s $
Show Answer
Answer:
Correct Answer: B
Solution:
Energy of electron in the 3rd orbit of $ H{e^{+}} $ is $ E _3=-13.6\times \frac{Z^{2}}{n^{2}}eV=-13.6\times \frac{4}{3^{2}}eV $
$ =-13.6\times \frac{4}{9}\times 1.6\times {10^{-19}}J $
From Bohr’s model, $ E _3=-KE _3=-\frac{1}{2}m _{e}v^{2} $
$ \Rightarrow \frac{1}{2}\times 9.1\times {10^{-31}}\times v^{2} $
$ =-13.6\times \frac{4}{9}\times 1.6\times {10^{-19}} $
$ \Rightarrow v^{2}=\frac{136\times 16\times 4\times 2\times {10^{-11}}}{9\times 91} $ or $ v=1.46\times 10^{6}m/s $
