Chapter 32, Problem 2P

To determine

Wavelength of 28GeV electron

Answer to Problem 2P

Solution:

The wavelength of 28GeV electron is $0.0\text{4}\times \text{1}{0}^{-\text{15}}\text{m or }0.0\text{4fm}.$

Explanation of Solution

Given Info:

$E=\text{energy of electron }=\text{28GeV}=\text{4}.\text{48}\times \text{1}{0}^{-\text{1}0}\text{J}$

Formula used:

The equation to determine the wavelength (l)of the electron is given by $E=\frac{hc}{\lambda }$ …………..I $\begin{array}{l}h=\text{planks constant }=\text{6}.\text{63}\times \text{1}{0}^{-\text{34}}\text{Js }\\ c=\text{velocity of light}=\text{3}\times \text{1}{0}^{\text{8}}\text{m}/\text{s}\\ \text{1eV}=\text{1}.\text{6x1}{0}^{-\text{19}}\text{J}.\end{array}$

Calculation:

Now rearranging equation I for l

$\lambda =\frac{hc}{E}$ ………………………………II

Substituting the values in equation II

We get,

$\lambda =(\text{6}.\text{63}\times \text{1}{0}^{-\text{34}}\text{Js })(\text{3}\times \text{1}{0}^{\text{8}}\text{m}/\text{s})/\text{4}.\text{48}\times \text{1}{0}^{-\text{1}0}\text{J}$

$\lambda =0.0\text{4}\times \text{1}{0}^{-\text{15}}\text{m or }0.0\text{4 fm}$