An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de-Broglie wavelength associated with the electrons. Taking other factors, such as numerical aperture etc. to be same, how does the resolving power of an electron microscope compare with that of an optical microscope which used yellow light?
The de-Broglie wavelength of the electrons is given by:
m = mass of the electron = 9.1×10-31 kg
e = charge on the electron = 1.6×10-19 C
V = accelerating potential = 50 kV
h = Planck's constant = 6.626×10-34 Js
Resolving power of a microscope, `R=(2musintheta)/lambda`
This formula suggests that to improve resolution, we have to use shorter wavelength and media with large indices of refraction. For an electron microscope, μ is equal to 1(vacuum).
For an electron microscope, the electrons are accelerated through a 60,000 V potential difference. Thus the wavelength of electrons is found to be
As, λ is very small (approximately 10-5 times smaller) for electron microscope than an optical microscope which uses yellow light of wavelength (5700 Å to 5900 Å). Hence, the resolving power of an electron microscope is much greater than that of optical microscope.
Electrons are accelerated by a voltage, V = 50 kV = 50 × 103 V
Charge on an electron, e = 1.6 × 10−19 C
Mass of an electron, me = 9.11 × 10−31 kg
Wavelength of yellow light = 5.9 × 10−7 m
The kinetic energy of the electron is given as:
E = eV
= 1.6 × 10−19 × 50 × 103
= 8 × 10−15 J
De Broglie wavelength is given by the relation:
`lambda = h/sqrt(2m_eE)`
`= (6.6 xx 10^(-34))/sqrt(2 xx 9.11 xx 10^(-31) xx 8 xx 10^(-15))`
`= 5.467 xx 10^(-12) m`
This wavelength is nearly 105 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 105 times that of an optical microscope.