Chapter 27, Problem 40P

To determine

The ratio of kinetic energy of an electron to kinetic energy of a proton.

Answer to Problem 40P

Solution:

The ratio of kinetic energy of an electron to that of a proton is $1.84\times {10}^3$

Explanation of Solution

Given:

Wavelength of the electron and proton is equal.

Formula Used:

$\begin{array}{l}\lambda =\frac{h}{p}\\ \text{KE}\text{=}\frac{1}{2}m{v}^2\end{array}$

Calculation:

By using the formula $\text{KE}\text{=}\frac{1}{2}m{v}^2$

The ratio of kinetic energy of electron to that of protons is given as

$\frac{{(KE)}_e}{{(KE)}_p}=\frac{\frac{1}{2}{m}_e{v}_e^2}{\frac{1}{2}{m}_p{v}_p^2}=\frac{m_e{v}_e^2}{m_p{v}_p^2}=\left(\frac{m_e}{m_p}\right)\left(\frac{v_e^2}{v_p^2}\right)$……………………………………………………… (1)

Now, according to the formula $\lambda =\frac{h}{p}$

Wavelength for electron and proton can be given as

${\lambda }_e=\frac{h}{m_e{v}_e}\text{and}{\lambda }_p=\frac{h}{m_p{v}_p}$

But the wavelength of the electron and proton is equal, that is

$\begin{array}{c}{\lambda }_e={\lambda }_p\\ \frac{h}{m_e{v}_e}=\frac{h}{m_p{v}_p}\\ \frac{m_p}{m_e}=\frac{v_e}{v_p}\end{array}$

Now on substituting the value $\frac{v_e}{v_p}$ in equation (1), we get

$\begin{array}{l}\frac{{(KE)}_e}{{(KE)}_p}=\left(\frac{m_e}{m_p}\right)\left(\frac{m_p^2}{m_e^2}\right)\\ \frac{{(KE)}_e}{{(KE)}_p}=\left(\frac{m_p}{m_e}\right)=\left(\frac{1.67\times {10}^{-27}\text{kg}}{9.11\times {10}^{-31}\text{kg}}\right)=1.84\times {10}^3\end{array}$

The ratio of kinetic energy of an electron to that of a proton is $1.84\times {10}^3$.