A) -2
B) \[-\frac{1}{2}\]
C) 2
D) \[\frac{1}{2}\]
Correct Answer: B
Solution :
We know that in any Bohr orbit of hydrogen atom Kinetic energy \[K.E.=\frac{1}{2}\left[ \frac{KZ{{e}^{2}}}{r} \right]\] ?.(i) Potential energy \[PE=-\left[ \frac{KZ_{e}^{2}}{r} \right]\] ??.(ii) Now, dividing equation (i) by (ii) we obtain \[\frac{KE}{PE}=\frac{\frac{1}{2}\left( \frac{KZ{{e}^{2}}}{r} \right)}{\left( \frac{KZ{{e}^{2}}}{r} \right)}=-\frac{1}{2}\]
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