Atomic StructureHard
Question
As the orbit number increases, the K.E. and P.E. for an electron
Options
A.both increases.
B.both decreases.
C.K.E. increases but P.E. decreases.
D.P.E. increases but K.E. decreases.
Solution
$K.E. = \frac{1}{2}mv^{2},P.E. = - mv^{2}\text{ and }V \propto \frac{1}{n}$
Therefore, with increase in orbit number, K.E. decreased but P.E. increases.
Create a free account to view solution
View Solution FreeMore Atomic Structure Questions
For hydrogen atom, the number of revolutions of the electron per second in the orbit of quantum number, n, is proportion..., The compound with the above configuration is named as :-...The dye acriflavine when dissolved in water has its maximum light absorption at 4530 Å and has maximum florescence emiss...The number of nodal planes in 2px orbital is...The formation of O2-(g) starting from O(g) is endothermic by 639 kJ mol-1 . If electron gain enthalpy of O(g) is –141 kJ...