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Question
An electron in an atom is revolving around the nucleus in a circular orbit of a radius of 5.3 x 10-11 m, with a speed of 2 x 106 m/s. Find the resultant orbital magnetic moment and angular momentum of the electron. [e = 1.6 x 10-19 C, me= 9.1 x 10-31 kg]
Solution 1
- Data: r = 5.3 × 10-11 m, v = 2 × 106 m/s,
e = 1.6 × 10-19 C, me = 9.1 × 10-31 kg
The orbital magnetic moment of the electron is
`"M"_0 = 1/2`evr
`= 1/2(1.6 xx 10^-19)(2xx10^6)(5.3 xx 10^-11)`
= 8.48 × 10-24 A.m2
The angular momentum of the electron is
L0 = mevr
= (9.1 × 10-31)(2 × 106)(5.3 × 10-11)
= 96.46 × 10-36 = 9.646 × 10-35 kg.m2/s
Solution 2
Given:
r = `5.3 xx 10^-11` m,
v = `2 xx 10^6` ms-1,
e = `1.6 xx 10^-19` C,
me = `9.1 xx 10^-31` kg
To find:
- Orbital magnetic moment (morb)
- Angular momentum of electron
Formulae:
- morb = `"evr"/2`
- L = mvr
Calculation:
From formula (i),
morb = `(1.6 xx 10^-19 xx 2 xx 10^6 xx 5.3 xx 10^-11)/2`
= `1.6 xx 5.3 xx 10^-24`
= `8.48 xx 10^-24` Am2
From formula (ii),
L = `9.1 xx 10^-31 xx 2 xx 10^6 xx 5.3 xx 10^-11`
= `96.46 xx 10^-36`
∴ L ≈ `underline(9.646 xx 10^-35)` kgm2/s
- Orbital magnetic moment is `8.48 xx 10^-24` Am2.
- Angular momentum of electron is `9.646 xx 10^-35` kgm2/s.
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