Question

Q6] The velocity of electron in the first Bohr's orbit of radius 0.5 A.U. is 2.24x10
${10}^{6}$
m/s. Calculate the period of revolution of the electron in the same orbit..​

Answer 1

Answer:

$mvr = \frac{nh}{2\pi} . \\ v = \frac{nh}{2\pi \: mr} . \\ = \frac{1 \times 6.63 \times 10 {}^{ - 34} }{2 \times 3.14 \times 9.1 \times 10 {}^{ - 31 \times 0.529 \times 10 {}^{ - 10} } } \\ = 2.187691 \times 10 {}^{6} ms {}^{ - 1} .$

Answer 2

Answer:

Period of revolution of the electron in same orbit is  1.4 × $10^{-16}$ s.

Explanation:

• According to Bohr model electrons are revolving in circular orbits around the nucleus. These orbits are called Bohr's orbit.
• If an electron of mass 'm' revolves with 'v' velocity along 'n' orbit of radius 'r' then for that electron, $mvr = \frac{nh}{2\pi }$
• Period of revolution: Time taken to complete one revolution around the nucleus is called period of revolution. It is expressed by the formula, T = $\frac{2\pi r}{v}$
• Given data: r = 0.5 A = 0.5 × $10^{-10}$ m [as 1 A =  $10^{-10}$ m]

                           v =  2.24 ×  $10^{6}$ m/s

         T = $\frac{2\pi \times0.5\times10^{-10}}{2.24\times10^6}$ = 1.4 × $10^{-16}$ s

        ∴ Period of revolution of the electron in same orbit is  1.4 × $10^{-16}$ s.

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