Chapter 27.2, Problem 28PP

To determine

To Find: The kinetic energy of proton.

Answer to Problem 28PP

  $4.19\times {10}^{-2}\text{eV}$

Explanation of Solution

Given data:

An electron with de Broglie wavelength of $0.14\text{ nm}$

Wavelength of electron $=$ wavelength of proton i.e. $0\cdot 14\text{ nm}$

Formula used:

de Broglie wavelength of proton is related to its velocity by:

  $\lambda =\frac{h}{mv}$

Here, h is the Planck’s constant, m is the mass of proton and v is the velocity of the proton.

The kinetic energy of the proton can be obtained by:

  $K.E=\frac{1}{2}m{v}^2$

Calculation:

  $\begin{array}{l}\lambda =\frac{h}{mv}\\ v=\frac{h}{m\lambda }\\ K.E=\frac{1}{2}m{\left(\frac{h}{m\lambda }\right)}^2\\ K.E=\frac{h^2}{2m{\lambda }^2}\end{array}$

Substitute all the known values and solve.

  $\begin{array}{l}K.E.=\frac{h^2}{2m{\lambda }^2}\\ K.E.=\frac{{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{2\times 1.67\times {10}^{-27}\text{kg}\times {\left(0.14\times {10}^{-9}\text{m}\right)}^2}\\ K.E.=6.71\times {10}^{-21}\text{J}\\ K.E.=\frac{6.71\times {10}^{-21}\text{J}}{1.6\times {10}^{-19}}=4.19\times {10}^{-2}\text{eV}\end{array}$

Conclusion:

Thus, the kinetic energy of the proton is $4.19\times {10}^{-2}\text{eV}$ .