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Expert-verified Found in: Page 713 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # A positron (charge +e, mass equal to the electron mass) is moving at 1.0x107 m/s in the positive direction of an x axis when, at x=0, it encounters an electric field directed along the x axis. The electric potential V associated with the field is given in Fig. 24-57. The scale of the vertical axis is set by Vs=500.0 V. (a) Does the positron emerge from the field at x=0 (which means its motion is reversed) or at x=0.50 m (which means its motion is not reversed)? (b) What is its speed when it emerges? 1. The positron emerges from the field at x=0, which means its motion is reversed.
2. The speed of the positron when it emerges is $1.0×{10}^{7}\mathrm{m}/\mathrm{s}$.
See the step by step solution

## Step 1: The given data

1. The speed of the positron at x=0, $\mathrm{v}=1.0×{10}^{7}\mathrm{m}/\mathrm{s}$
2. The mass of the positron, $\mathrm{m}=9.1×{10}^{-31}\mathrm{kg}$
3. The scale of the vertical axis, data-custom-editor="chemistry" ${\mathrm{V}}_{\mathrm{s}}=500\mathrm{V}$

## Step 2: Understanding the concept of energy

Using the concept of energy, we can get the kinetic energy of the positron. This will help in determining the potential height or energy of the system and hence, we can check the emerging point of the positron. Now, comparing the concept with the given situation, we can get the speed of the positron.

Formulae:

The potential energy of the system in terms of potential difference and charge is given by, U=q(deltaV) (i)

The kinetic energy of the system, $\mathrm{K}.\mathrm{E}=\frac{1}{2}{\mathrm{mv}}^{2}$ (ii)

## Step 3: a) Calculation of the position at which the positron emerges

The initial kinetic energy of the positron at x = 0 is given using equation (ii) as given:

$\begin{array}{rcl}\mathrm{KE}& =& \frac{1}{2}\left(9.11×{10}^{-31}\mathrm{kg}\right){\left(1.0×{10}^{7}\mathrm{m}/\mathrm{s}\right)}^{2}\\ & =& 4.55×{10}^{-17}\mathrm{J}\\ & =& \left(4.55×{10}^{-17}\mathrm{J}\right)\left(\frac{1\mathrm{eV}}{1.6×{10}^{-19}\mathrm{J}}\right)\\ & =& 284.37\mathrm{eV}\\ & \approx & 284eV\end{array}$

The height of the potential energy barrier is given using equation (i) as follows:

data-custom-editor="chemistry" $\begin{array}{rcl}\mathrm{U}& =& \mathrm{e}\left(500\mathrm{V}\right)\\ & =& 500\mathrm{eV}\end{array}$

The height of the potential energy barrier is greater as compared with the initial kinetic energy of the positron.

So, the positron emerges from the barrier at x=0, which means its motion is reversed.

## Step 4: b) Calculation of the speed of the positron

When the positron emerges from the field, the positron is moving in the opposite direction.

So, the final velocity of the positron is directed along the negative x-direction but its magnitude does not change.

Therefore, the speed of the positron is $1.0×{10}^{7}\mathrm{m}/\mathrm{s}$ ### Want to see more solutions like these? 