Chapter 27, Problem 43P

To determine

The proton has shorter wavelength, if electron and proton have same kinetic energy.

Answer to Problem 43P

Solution:

When electron and proton have same kinetic energy, the wavelength of proton is shorter.

Explanation of Solution

Given:

Kinetic Energy 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 $\text{KE}\text{=}\frac{1}{2}m{v}^2=\frac{p^2}{2m}$

$p={\left[2m(\text{KE})\right]}^{\frac{1}{2}}$

The wavelength can be given as $\lambda =\frac{h}{p}$

So, $\lambda =\frac{h}{{\left[2m(\text{KE})\right]}^{\frac{1}{2}}}$

If we take the ratio of wavelength of electron and proton with same kinetic energy, we will find that

$\begin{array}{l}\frac{{\lambda }_e}{{\lambda }_p}=\frac{\frac{h}{{\left[2m_e(\text{KE})\right]}^{\frac{1}{2}}}}{\frac{h}{{\left[2m_p(\text{KE})\right]}^{\frac{1}{2}}}}={\left(\frac{m_p}{m_e}\right)}^{\frac{1}{2}}\\ \frac{{\lambda }_e}{{\lambda }_p}={\left(\frac{m_p}{m_e}\right)}^{\frac{1}{2}}\end{array}$

By the above relation, as the $m_p>m_e\text{so}{\lambda }_p<{\lambda }_e$

Thus we have derived that for same kinetic energy; the wavelength of proton is shorter.