Question
The $$L{i^{2 + }}\,ion$$ is moving in the third stationary state, and its linear momentum is $$7.3 \times {10^{ - 34}}kgm{s^{ - 1}}.$$ Angular momentum is.
A.
$$1.158 \times {10^{ - 45}}kg\,{m^2}{s^{ - 1}}$$
B.
$$11.58 \times {10^{ - 48}}kg\,{m^2}{s^{ - 1}}$$
C.
$$11.58 \times {10^{ - 47}}kg\,{m^2}{s^{ - 1}}$$
D.
$$12 \times {10^{ - 45}}kg\,{m^2}{s^{ - 1}}$$
Answer :
$$11.58 \times {10^{ - 48}}kg\,{m^2}{s^{ - 1}}$$
Solution :
$$\eqalign{
& Z = 3\,\,{\text{for}}\,L{i^{2 + }}\,ions \cr
& {\text{So}}\,\,\,{r_n} = \frac{{52.9 \times {n^2}}}{Z}pm \cr
& n = 3,Z = 3 \cr
& {r_n} = \frac{{52.9 \times {{\left( 3 \right)}^2}}}{3}pm \cr
& = 158.7\,pm \cr
& {\text{Also, linear momentum}}\,\left( {mv} \right) \cr
& = 7.3 \times {10^{ - 34}}kg\,m{s^{ - 1}} \cr
& {\text{Then angular momentum will be}} \cr
& \omega = \left( {mv} \right) \times r \cr
& = \left( {7.3 \times {{10}^{ - 34}}kg\,m{s^{ - 1}}} \right)\left( {158.7\,pm} \right) \cr
& = 7.3 \times {10^{ - 34}}kg\,m{s^{ - 1}} \times \left( {158.7 \times {{10}^{ - 12}}m} \right) \cr
& = 11.58 \times {10^{ - 48}}kg\,{m^2}{s^{ - 1}} \cr} $$