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)You need to login to perform this action.
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