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Question
Calculate the change in angular momentum of the electron when it jumps from third orbit to first orbit in hydrogen atom.
(Take h = 6.33 × 10−34 Js)
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Solution
According to Bohr's second postulate,
Angular momentum = `(nh)/(2 pi)`
For the first orbit, n1 = 1
∴ `L_1 = ((1)h)/(2 pi) = h/(2 pi)` ...(i)
For the third orbit, n3 = 3
∴ `L_3 = ((3)h)/(2 pi) = (3h)/(2 pi)` ...(ii)
When an electron jumps from 3rd orbit to 1st orbit, the change in angular momentum is
`L_3 - L_1 = (3 h)/(2 pi) - h/(2 pi)`
= `(2 h)/(2 pi)`
= `h / pi`
Putting h and π values, we get
Change in angular momentum = `(6.33 xx 10^-34)/3.142`
= 2.11 × 10−34 kg m2/s
The change in angular momentum of an electron when it jumps from the 3rd orbit to the 1st orbit in a hydrogen atom is 2.11 × 10−34 kg m2/s.
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