question_answer

A doubly ionized \[\text{L}{{\text{i}}^{\text{2+}}}\] ion in ground state absorbs 91.8 eV of energy. Find the increase in angular momentum of electron. (Take \[h=6.63\times {{10}^{-34}}J-s\])

A)  \[\text{2}\text{.11 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-34}}}\text{J-s}\]       

B)  \[\text{3}\text{.16 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-34}}}\text{J-s}\]

C)  \[\text{1}\text{.05 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-34}}}\text{J-s}\]       

D)  \[\text{4}\text{.22 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-34}}}\text{J-s}\]

Correct Answer: C

Solution :

 When 91.8 eV energy has been absorbed, electron transits in n = 2 state. Change in angular momentum \[=\frac{2h}{2\pi }-\frac{h}{2\pi }\] \[=\frac{h}{2\pi }\] \[=1.05\times {{10}^{-34}}\text{J-s}\]