Question

(a) Find the wavelength of a proton whose kinetic energy is equal 10 its integral energy.

(b) ' The proton is usually regarded as being roughly of radius${\mathbf{10}}^{\mathbf{-}\mathbf{15}}\mathbf{m}$. Would this proton behave as a wave or as a particle?

Short answer

(a)The wavelength of a proton that has kinetic energy equal to its internal energy is $7.63\times {10}^{-16}\mathrm{m}$

(b) The moving proton would behave like as a particle in nature.

Step-by-step solution

Given data

$$\begin{gathered} \mathrm{Mass}\mathrm{of}\mathrm{proton},m=1.67\times {10}^{-27}\mathrm{kg} \\ \mathrm{Speed}\mathrm{of}\mathrm{light},c=3.0\times {10}^8\mathrm{m}/\mathrm{s} \\ \mathrm{Plank}\text{'}\mathrm{s}\mathrm{constant},h=6.63\times {10}^{-34}\mathrm{J}.\mathrm{s} \end{gathered}$$

Relativistic effect in de Broglie's equation

The internal energy Eof an object is

$\mathit{E}\mathbf{=}\mathit{m}{\mathit{c}}^{\mathbf{2}}$

The equation for the kinetic energy ( KE)of an object traveling at relativistic velocities is

$\mathit{K}\mathit{E}\mathbf{=}\mathbf{(}{\mathit{\gamma }}_{\mathbf{u}}\mathbf{-}\mathbf{1}\mathbf{)}\mathit{m}{\mathit{c}}^{\mathbf{2}}$

Where, mis rest mass, and cis the speed of light.

De Broglie's wavelength

$\mathit{\lambda }\mathbf{=}\frac{\mathbf{h}}{\mathbf{p}}$

Where, pis the momentum.

Relativistic effect in de Broglie's equation

$\mathit{\lambda }\mathbf{=}\frac{\mathbf{h}}{{\mathbf{\gamma }}_{\mathbf{u}}\mathbf{m}\mathbf{v}}$

Where, is the velocity at which the protons kinetic energy equals its internal energy, and$\mathit{\gamma }$is Lorentz factor.

Speed at which the internal energy become equal to the kinetic energy

$$\begin{gathered} \mathit{m}{\mathit{c}}^{\mathbf{2}}\mathbf{=}\left({\gamma }_u-1\right) \\ \mathbf{1}\mathbf{=}{\mathit{\gamma }}_{\mathbf{u}}\mathbf{-}\mathbf{1} \\ \mathbf{2}\mathbf{=}{\mathit{\gamma }}_{\mathbf{u}} \end{gathered}$$

Lorentz factor is given by a relation,

$$\begin{gathered} {\mathit{\gamma }}_{\mathbf{u}}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}{\displaystyle \frac{{\mathbf{v}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}}} \\ \mathbf{}\mathbf{}\mathbf{2}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}{\displaystyle \frac{v^2}{c^2}}}} \\ \sqrt{\mathbf{1}\mathbf{-}\frac{{\mathbf{v}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{1}\mathbf{-}\frac{{\mathbf{v}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\frac{{\mathbf{v}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{3}}{\mathbf{4}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{v}\mathbf{=}\frac{\sqrt{\mathbf{3}}}{\mathbf{2}}\mathit{c}\mathbf{} \end{gathered}$$

Substitute the value of velocity in wavelength, 

(a)

Substitute the value of velocity in wavelength $\lambda$,

The wavelength of a proton that has kinetic energy equal to its internal energy is given by,

$\lambda =\frac{h}{\gamma m\left({\displaystyle \frac{\sqrt{3c}}{2}}\right)}$

Substitute $6.63\times {10}^{-34}\mathrm{J}.\mathrm{s}\mathrm{for}\mathrm{h},1.67\times {10}^{-27}\mathrm{kg}\mathrm{for}\mathrm{m}\mathrm{and}3.0\times {10}^8\mathrm{m}/\mathrm{s}$for c in the above equation to solve for $\lambda$

$$\begin{gathered} \lambda =\frac{6.63\times {10}^{-34}J.s}{\left(2\right)(1.67\times {10}^{-27}\mathrm{kg})(\sqrt{{\displaystyle \frac{3\times 3.0\times {10}^8\mathrm{m}/\mathrm{s}}{2}}}} \\ =7.63\times {10}^{-16}\mathrm{m} \end{gathered}$$

Hence, the wavelength of a proton that has kinetic energy equal to its internal energy is$\lambda =7.63\times {10}^{-16}\mathrm{m}$

Explain proton behavior

(b)

The wavelength of a proton that has kinetic energy equal to its internal energy is$\mathrm{\lambda }=7.63\times {10}^{-16}\mathrm{m}$

As the wavelength of the proton is smaller than the roughly size of the proton.

$\left({10}^{-15}\mathrm{m}\right)$So, the moving proton would behave like as a particle in nature.