Question

Show that the units of$\frac{{\mathbf{P}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}$ and$\mathit{m}{\mathit{c}}^{\mathbf{2}}$are indeed joules.

Short answer

Both the terms$\frac{p^2}{2m}$ and$m{c}^2$ have the same unit, which is Joule

Step-by-step solution

Formula of energy

The particle’s energy is given by:

$\mathit{E}\mathbf{=}\frac{{\mathbf{p}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}$

Here,is the momentum of the particle andis the mass of the particle inkg.

The momentum is given by:

Here is mass, and is the velocity of the particle.

Evaluate the first equation

Use the formula of momentum into the formula of energy.

$$\begin{gathered} E=\frac{{\left(mv\right)}^2}{2m} \\ =\frac{1}{2}m{v}^2 \end{gathered}$$

Therefore, it is safe to say that$\frac{p^2}{2m}$ is equal to $\frac{1}{2}m{v}^2$.

Finding the unit of p22m

Find the unit of the quantity $\frac{p^2}{2m}$and use $\left(kg\right)\left(m/s^2\right)=N$,

$$\begin{gathered} \left(1kg\right){\left(1m/s^2\right)}^2=\left(1kg\right)\left(1m/s\right)\left(1m\right) \\ =\left(1kg\right)\left(1m/s^2\right)\left(1m\right)\times \frac{1N}{1kgm/s} \\ =1Nm\times \frac{1J}{1Nm} \\ =1J \end{gathered}$$

Finding the unit of mc2

The formula$m{c}^2$presents the rest energy of the particle where c is the velocity of light and its unit is$m/s$.

Find the unit of the quantity mc² and use$\left(kg\right)\left(m/s^2\right)=N$.

$$\begin{gathered} \left(1kg\right)\left(1m/s\right)=\left(1kg\right)\left(1m/s\right)\left(1m\right) \\ =\left(1kg\right)\left(1m/s^2\right)\left(1m\right)\times \frac{1N}{1kgm/s} \\ =1Nm\times \frac{1J}{1Nm} \\ =1J \end{gathered}$$

Therefore, both the terms$\frac{p^2}{2m}$ and mc² have the same unit, which is .