Atoms And Nuclei Question 288
Question: Consider 3rd 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 and h (Plank’s Constant) $ 6.6\times {{10}^{-34}}Js $ ]
Options:
A) $ 1.46\times 10^{6}m/s $
B) $ 0.73\times 10^{6}m/s $
C) $ 3.0\times 10^{8}m/s $
D) $ 2.92\times 10^{6}m/s $
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Answer:
Correct Answer: A
Solution:
-
Speed of electron in nth orbit $ V_{n}=\frac{2\pi KZe^{2}}{nh} $
$ V=( 2.19\times 10^{6}m/s )\frac{Z}{n} $ $ V=(2.19\times 10^{6})\frac{2}{3}( Z=2\And n=3 ) $
$ V=1.46\times 10^{6}m/s $
