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 $

Show Answer

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 $



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