Modern Physics 7 Question 26
27. In an electron microscope, the resolution that can be achieved is of the order of wavelength of electrons used. To resolve a width of $7.5 \times 10^{-12} m$, the minimum electron energy required is close to
(Main 2019, 10 Jan I)
(a) $500 keV$
(b) $1 keV$
(c) $100 keV$
(d) $25 keV$
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Solution:
- Given, resolution achieved in electron microscope is of the order of wavelength.
So, to resolve $7.5 \times 10^{-12} m$ seperation wavelength associated with electrons is
$$ \lambda=7.5 \times 10^{-12} m $$
$\therefore$ Momentum of electrons required is
$$ p=\frac{h}{\lambda} $$
or kinetic energy of electron must be
$$ KE=\frac{p^{2}}{2 m}=\frac{(h / \lambda)^{2}}{2 m} $$
Substituting the given values, we get
$$ \begin{aligned} & =\frac{\frac{6.6 \times 10^{-34}}{7.5 \times 10^{-12}}}{2 \times 9.1 \times 10^{-31}} J \\ & =\frac{\left(6.6 \times 10^{-34}\right)^{2}}{2 \times 9.1 \times 10^{-31} \times\left(7.5 \times 10^{-12}\right)^{2} \times\left(1.6 \times 10^{-19}\right)} eV \\ & \quad\left(\because 1 eV=1.6 \times 10^{-19} J\right) \\ & =26593.4 \simeq 26.6 \times 10^{3} eV \simeq 26 keV \end{aligned} $$
which is nearest to $25 keV$.
