A) KE and PE both increases
B) KE and PE both decreases
C) KE increases while PE decreases
D) KE decreases while PE increases
Correct Answer: D
Solution :
For a hydrogen-like atom, \[TE=-\frac{Z{{e}^{2}}}{4\pi {{\varepsilon }_{0}}\times 2r}\] \[KE=+\frac{Z{{e}^{2}}}{8\pi {{\varepsilon }_{0}}r}\] \[PE=-\frac{Z{{e}^{2}}}{4\pi {{\varepsilon }_{0}}r}\] As electron moves from lower energy state to higher energy state, its TE increases, and hence KE decrease and PE becomes less negative, i.e., it increases. (Note that speed also decreases)
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