Question

Calculate ratio of velocity for he+ ion in 2nd excited state to the Li+2 ion in 3rd excited state...
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Answer 1

Answer: velocity=2.18×10^6(z/n)

Hence the ratio =2×3/3×4=1/2

Ratio is 1:2

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Answer 2

Explanation:

2nd excited state, means e− is present in 3rd shell of hydrogen

2nd excited state, means e− is present in 3rd shell of hydrogenr3=0.529×(3)21=0.529×9

2nd excited state, means e− is present in 3rd shell of hydrogenr3=0.529×(3)21=0.529×91st excited state, means e− exist in 2nd shell of Li+2

r2=0.529×(2)23

r2=0.529×(2)23=0.529×43⇒(r3)H(r2)Li+2=0.529×910.529×43

r2=0.529×(2)23=0.529×43⇒(r3)H(r2)Li+2=0.529×910.529×43=radius of 2nd excited state of hydrogenradius of 1st excited state of Li+2⇒(r3)H(r2)Li+2=274

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