Respuesta :
Explanation:
(a) The given data is as follows.
Speed of electron (u) = [tex]5.5 \times 10^{4} m/s[/tex]
According to De Broglie's formula,
wavelength, [tex]\lambda = \frac{h}{mu}[/tex]
where, h = Planck's constant = [tex]6.626 \times 10^{-34} Js[/tex]
m = mass of electron = [tex]9.11 \times 10^{-31} kg[/tex]
Hence, we will calculate the wavelength as follows.
[tex]\lambda = \frac{h}{mu}[/tex]
= [tex]\frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31}kg \times 5.5 \times 10^{4} m/s}[/tex]
= [tex]0.132 \times 10^{-7}[/tex] m
= [tex]13.2 \times 10^{-9}[/tex] m
It is known that for any microscope, smallest object that can be observed is equal to [tex]\frac{1}{2}[/tex] the wavelength of of the radiation smallest object observable with an electron microscope.
Hence, [tex]\frac{13.2 \times 10^{-9}}{2}[/tex]
= [tex]6.6 \times 10^{-9}[/tex] m
= 6.6 nm (as 1 m = [tex]10^{-9} nm[/tex])
Therefore, the smallest object observable with an electron microscope will be 6.6 nm.
(b) At [tex]3.0 \times 10^{7} m/s[/tex], the wavelength will be calculated as follows.
wavelength, [tex]\lambda = \frac{h}{mu}[/tex]
= [tex]\frac{6.626 \times 10^{-34} Js}{9.11 \times 10^{-31} kg \times 3.0 \times 10^{7} m/s}[/tex]
= [tex]24.2 \times 10^{-12}[/tex] m
As, for any microscope, smallest object that can be observed is equal to [tex]\frac{1}{2}[/tex] the wavelength of of the radiation smallest object observable with an electron microscope.
= [tex]\frac{24.2 \times 10^{-12}}{2}[/tex]
= [tex]12.1 \times 10^{-12} m \times 10^{9}nm/m[/tex]
= 0.0121 nm
Therefore, at [tex]3.0 \times 10^{7} m/s[/tex] the smallest object observable with an electron microscope is 0.0121 nm.
