An electron microscope uses electrons accelerated by a voltage of `50kV`. Determine the De Broglie wavelength associated with the electrons. If other factors ( such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?

Explanation

Electrons are acceleratred by a voltage , `V=50 kV =50xx10^(3) V`
Charge on an electron , `e=1.6xx10^(-19) C`
Mass of an electron , `m_e=9.11xx10^(-31) kg`
Wavelength of yellow light =`5.9xx10^(-7)` m
The kineti energy of the electron is given as : E=eV
`=1.6xx10^(-19)xx50xx10^3`
`=8xx10^(-15) J`
De Broglie wavelength is given by the relation :
`lambda=h/sqrt(2mE)`
`=(6.6xx10^(-34))/sqrt(2xx9.11xx10^(-31)xx8xx10^(-15))`
`=5.467xx10^(-12)` m
This wavelength is nearly `10^5` times less than the wavelength of yellow light. The resolving power of a microscope is inversely proportional to the wavelength of light used. Thus , the resolving power of an electron microscope is nearly `10^5` times that of an optical microscope.