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Expert-verified Found in: Page 733 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # Electron guns are used in X-ray tubes. The electrons are accelerated through a relatively large voltage and directed onto a metal target, producing X-rays. (a) How many electrons per second strike the target if the current is 0.500 mA ? (b) What charge strikes the target in 0.750 s?

(a) About $$3.13 \times {10^{15}}\;{\rm{electron}}$$per second strike the target.

(b) The target has been stroked with a charge of $${\rm{3}}{\rm{.75 \times 1}}{{\rm{0}}^{{\rm{ - 4}}}}\;{\rm{C}}$$.

See the step by step solution

## Step 1: Identification of the given data

The given data can be listed below as:

• The value of the current is $$I = 0.500\;{\rm{mA \times }}\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{A}}}}{{{\rm{1}}\;{\rm{mA}}}}{\rm{ = 0}}{\rm{.5 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{A}}$$.
• The time taken to strike the target is t = 0.750 s.

## Step 2: Significance of the electric charge

The electric charge is described as the charge carried by an electron. The electric charge is described as the product of the intensity and the time taken to reach to the current.

## Step 3: (a) Determination of the number of electrons

The equation of the number of electrons is expressed as:

$$N = \frac{{I{t_1}}}{{{Q_e}}}$$

Here, $$N$$ is the number of electrons, l is the value of the current, $${t_1}$$ is the time taken to strike the target that is one second and $${Q_e}$$ is the charge of one electron.

Substitute the values in the above equation.

\begin{aligned}N &= \frac{{\left( {0.5 \times {{10}^{ - 3}}\,{\rm{A}}} \right)\left( {1\;{\rm{s}}} \right)}}{{1.6 \times {{10}^{ - 19}}\;{\rm{C/electron}}}}\\ & = \frac{{\left( {0.5 \times {{10}^{ - 3}}\,{\rm{A}} \cdot {\rm{s}}} \right)}}{{1.6 \times {{10}^{ - 19}}\;{\rm{C/electron}}}}\\ & = \frac{{\left( {0.5 \times {{10}^{ - 3}}\,{\rm{A}} \cdot {\rm{s}} \times \frac{{1\;{\rm{C}}}}{{1\;{\rm{A}} \cdot {\rm{s}}}}} \right)}}{{1.6 \times {{10}^{ - 19}}\;{\rm{C/electron}}}}\\ & = 3.13 \times {10^{15}}\;{\rm{electron}}\end{aligned}

Thus, about $$3.13 \times {10^{15}}\;{\rm{electron}}$$ per second strike the target.

## Step 4: (b) Determination of the charge

The equation of the charge is expressed as:

$$Q = It$$

Here, $$Q$$ is the charge and t is the time taken to strike the target.

Substitute the values in the above equation.

\begin{aligned}Q = {\rm{0}}{\rm{.5 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{A}} \times 0.750\;{\rm{s}}\\{\rm{ = 3}}{\rm{.75 \times 1}}{{\rm{0}}^{{\rm{ - 4}}}}\;{\rm{A}} \cdot {\rm{s}}\\{\rm{ = 3}}{\rm{.75 \times 1}}{{\rm{0}}^{{\rm{ - 4}}}}\;{\rm{A}} \cdot {\rm{s}} \times \frac{{1\;{\rm{C}}}}{{1\;{\rm{A}} \cdot {\rm{s}}}}\\ = {\rm{3}}{\rm{.75 \times 1}}{{\rm{0}}^{{\rm{ - 4}}}}\;{\rm{C}}\end{aligned}

Thus, the target has been stroked with a charge of $${\rm{3}}{\rm{.75 \times 1}}{{\rm{0}}^{{\rm{ - 4}}}}\;{\rm{C}}$$. ### Want to see more solutions like these? 