A) \[{{l}_{H}}>{{l}_{Li}}\,\text{ and }|{{E}_{H}}|>|{{E}_{Li}}|\]
B) \[{{l}_{H}}={{l}_{Li}}\text{ and }|{{E}_{H}}|<|{{E}_{Li}}|\]
C) \[{{l}_{H}}={{l}_{Li}}\text{ and }|{{E}_{H}}|>|{{E}_{Li}}|\]
D) \[{{l}_{H}}<{{l}_{Li}}\text{ and }|{{E}_{H}}|<|{{E}_{Li}}|\]
Correct Answer: B
Solution :
In second excited state n = 3 So \[{{l}_{H}}={{l}_{Li}}=3\left( \frac{h}{2\pi } \right)\] While E µ Z2 and ZH = 1, ZLi = 3 So |ELi| = 9|EH| or |EH|<|ELi|You need to login to perform this action.
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