Calculate the wavelength of an electron that has been accelerated in a particle accelerator through a potential difference of 100 million volts. 

Explanation

The kinetic energy of an electron under the potential difference of 100 meV or `10^(8)V` is given by the equation, `(1)/(2)mv^(2)=eV`
Given, `e=1.6xx10^(-19)C, v=10^(8)V`
`(1)/(2)xx9.1xx10^(-31)xxv^(2)=1.6xx10^(-19)Cxx10^(8)V`
`therefore v^(2)=(2xx1.6xx10^(-19)xx10^(8))/(9.1xx10^(-31))=0.352xx10^(20)m^(2)s^(-2) or v=5.93xx10^(9)xx10^(9)ms^(-1)`
The de Broglie wavelength
`lambda=(h)/(mv)=(6.6xx10^(-34)kgm^(2)s^(-1))/(9.1xx10^(-31)kgxx5.93xx10^(9)m^(-1))=1.22xx10^(-13)m`