91Ó°ÊÓ

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" ${\mathrm{V}}_{\mathrm{s}}=500\mathrm{V}$

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)

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

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

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.

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 $1.0×{10}^7\mathrm{m}/\mathrm{s}$