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15 PE

Expert-verifiedFound in: Page 82

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**The planetary model of the atom pictures electrons orbiting the atomic nucleus much as planets orbit the Sun. In this model you can view hydrogen, the simplest atom, as having a single electron in a circular orbit **\({\bf{1}}{\bf{.06 \times 1}}{{\bf{0}}^{{\bf{ - 10}}}}\;{\bf{m}}\)** ****in diameter. (a) If the average speed of the electron in this orbit is known to be**\({\bf{2}}{\bf{.20 \times 1}}{{\bf{0}}^{\bf{6}}}{\bf{ m/s}}\)** ****, calculate the number of revolutions per second it makes about the nucleus. (b) What is the electron’s average velocity?**

(a) The number of revolutions per second about nucleus is \({\bf{6}}{\bf{.61 \times 1}}{{\bf{0}}^{{\bf{15}}}}\;{\bf{/s}}\).

(b) The average velocity of electron in circular orbit is \({\bf{0}}\;{\bf{m/s}}\).

(a)

Given Data:

The diameter of circular orbit is \(d = 1.06 \times {10^{ - 10}}\;{\rm{m}}\)

The average speed of electron in orbit is \(v = 2.20 \times {10^6}\;{\rm{m}}/{\rm{s}}\).

**The number of revolutions per second can be found by calculating the circumference of the circular orbit for one revolution and then divide the average speed by circumference of the circular orbit.**

The number of revolutions per second for electron in circular orbit is given as

\(v = \pi dN\)

Here, \(N\) is the number of revolutions per second for electron.

Substitute all the values in the above equation.

\(\begin{array}{c}2.20 \times {10^6}\;{\rm{m}}/{\rm{s}} = \pi \left( {1.06 \times {{10}^{ - 10}}\;{\rm{m}}} \right)N\\N = 6.61 \times {10^{15}}\;/{\rm{s}}\end{array}\)

Therefore, the number of revolutions per second about nucleus is \(6.61 \times {10^{15}}\;/{\rm{s}}\).

**(b)**

The average displacement of the electron in circular orbit is zero so the average velocity of electron in circular orbit is zero.

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