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Expert-verifiedFound in: Page 733

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**SPEAR, a storage ring about ****\({\bf{72}}.{\bf{0}}\;{\bf{m}}\) in diameter at the Stanford Linear Accelerator (closed in 2009), has a \({\bf{20}}.{\bf{0}} - {\bf{A}}\) circulating beam of electrons that are moving at nearly the speed of light. (See Figure 20.39****.) How many electrons are in the beam?**

About \(9.42 \times {10^{13}}\;{\rm{electron}}\) number of electrons are there.

** **

The given data can be listed below as:

- The diameter of the storage ring is \(d = 72.0\;{\rm{m}}\).
- The current carried by the circulating beam of electrons is \(I = 20.0\;{\rm{A}}\).

**The charge of a ring is described as the amount of current carried out by a ring in unit time. The charge carried is directly proportional to the current and the diameter of an object and inversely proportional to the velocity of that object.**

The equation of the number of electrons inside the beam is expressed as:

\(n = \frac{{\pi Id}}{{c{q_e}}}\)

Here, \(n\) is the number of electrons inside the beam, \(I\) is the current carried by the circulating beam of electrons, \(d\)is the diameter of the storage ring,\(c\) is the velocity of light and \({q_e}\) is the charge of one electron.

Substitute the values in the above equation.

\(\begin{align}n &= \frac{{\left( {3.14} \right)\left( {20.0\;{\rm{A}}} \right)\left( {72.0\;{\rm{m}}} \right)}}{{\left( {3 \times {{10}^8}\;{\rm{m/s}}} \right)\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C/electron}}} \right)}}\\ &= \frac{{\left( {62.8\;{\rm{A}}} \right)\left( {72.0\;{\rm{m}}} \right)}}{{\left( {3 \times {{10}^8}\;{\rm{m/s}}} \right)\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C/electron}}} \right)}}\\ &= \frac{{\left( {4521.6\;{\rm{A}} \cdot {\rm{m}}} \right)}}{{\left( {3 \times {{10}^8}\;{\rm{m/s}}} \right)\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C/electron}}} \right)}}\\ &= \frac{{\left( {1.507 \times {{10}^{ - 5}}\;{\rm{A}} \cdot {\rm{s}}} \right)}}{{\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C/electron}}} \right)}}\end{align}\)

Hence, further as:

\(\begin{align}n &= \frac{{\left( {1.507 \times {{10}^{ - 5}}\;{\rm{A}} \cdot {\rm{s}}} \right)}}{{\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C/electron}}} \right)}}\\ &= \frac{{\left( {1.507 \times {{10}^{ - 5}}\;{\rm{A}} \cdot {\rm{s}} \times \frac{{1\;{\rm{C}}}}{{1\;{\rm{A}} \cdot {\rm{s}}}}} \right)}}{{\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C/electron}}} \right)}}\\ &= \frac{{\left( {1.507 \times {{10}^{ - 5}}\;{\rm{C}}} \right)}}{{\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C/electron}}} \right)}}\\ &= 9.42 \times {10^{13}}\;{\rm{electron}}\end{align}\)

Thus, about \(9.42 \times {10^{13}}\;{\rm{electron}}\) number of electrons are

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