An electron microscope uses electrons accelerated by a voltage of 50 kV. 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?
Given: The electrons are accelerated by a voltage, 50 kV .
The kinetic energy of the electron is given as,
E = e V
Where, e is the charge of electron and V is the voltage.
By substituting the given values in the above equation, we get
E = 1.6 × 10 − 19 × 50 × 10 3 = 8 × 10 − 15 J
The de Broglie wavelength is given as,
λ = h 2 m e E
By substituting the value in the above expression, we get
λ = 6.6 × 10 − 34 2 × 9.1 × 10 − 31 × 8 × 10 − 15 = 5.467 × 10 − 12 m
Wavelength of yellow light is 5.9 × 10 − 7 m .
Thus, wavelength of an electron is nearly 10 5 times less than the wavelength of yellow light.
Since, the resolving power of a microscope is inversely proportional to the wavelength of light used, the resolving power of an electron microscope will be 10 5 times that of an optical microscope.
Given: The electrons are accelerated by a voltage,
The kinetic energy of the electron is given as,
Where,
By substituting the given values in the above equation, we get
The de Broglie wavelength is given as,
By substituting the value in the above expression, we get
Wavelength of yellow light is
Thus, wavelength of an electron is nearly
Since, the resolving power of a microscope is inversely proportional to the wavelength of light used, the resolving power of an electron microscope will be
