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Answer the following.
Write Bohr’s equations for circular motion of an electron around the nucleus and conservation of…
(a)[2.5]
Write Bohr’s equations for circular motion of an electron around the nucleus and conservation of angular momentum. Using them, obtain an expression for orbital velocity of an electron.
(b)[2.5]
Prove that $1\text{u} = 931\cdot5\text{ MeV}$ where the terms have their usual meaning.
To show: 1\text{u} = 931\cdot5\text{ MeV}
Answer
Answer (a)
AIDerivation of orbital velocity of an electron in Bohr's model:
Centripetal force: $\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r^2}$
Angular momentum: $mvr = \frac{nh}{2\pi} \implies \frac{1}{r} = \frac{2\pi mv}{nh}$
Substituting $\frac{1}{r}$: $mv^2 = \frac{Ze^2}{4\pi\epsilon_0}\left(\frac{2\pi mv}{nh}\right) \implies v = \frac{Ze^2}{2\epsilon_0 nh}$.
- Centripetal force is provided by electrostatic attraction: $\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{(Ze)(e)}{r^2}$.
- Bohr's quantization of angular momentum: $mvr = \frac{nh}{2\pi} \implies r = \frac{nh}{2\pi mv}$.
- Substitute $r$ into the centripetal equation: $mv^2 = \frac{Ze^2}{4\pi\epsilon_0} \left(\frac{2\pi mv}{nh}\right)$.
- Divide both sides by $mv$: $v = \frac{Ze^2}{2\epsilon_0 nh}$.
Answer (b)
Checked answer.
$1\text{ u} = 1\cdot6605 \times 10^{-27}$ kg. By $E = mc^2$: $E = 1\cdot6605 \times 10^{-27} \times (2\cdot998 \times 10^8)^2 = 1\cdot4924 \times 10^{-10}$ J. $E = \frac{1\cdot4924 \times 10^{-10}}{1\cdot602 \times 10^{-19}}$ eV $= 9\cdot315 \times 10^8$ eV $= 931\cdot5$ MeV.
- $1\text{ u} = \frac{1}{12}$ of the mass of a carbon-12 atom $= 1\cdot6605 \times 10^{-27}$ kg.
- Speed of light $c = 2\cdot998 \times 10^8$ m s$^{-1}$.
- By Einstein's mass-energy relation, $E = mc^2 = 1\cdot6605 \times 10^{-27} \times (2\cdot998 \times 10^8)^2 = 1\cdot4924 \times 10^{-10}$ J.
- $1$ eV $= 1\cdot602 \times 10^{-19}$ J, so $E = \frac{1\cdot4924 \times 10^{-10}}{1\cdot602 \times 10^{-19}}$ eV $= 9\cdot315 \times 10^8$ eV.
- $E = 931\cdot5$ MeV, so $1\text{ u} = 931\cdot5$ MeV.
From ISC 2027 Specimen Physics Paper 1, question 19(i).