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  • We wish to see inside an atom. Assuming the atom to have a diameter of 100 pm <img alt="\mathrm{\text { [1 picometer }(\mathrm{pm}), 10^{-12} \mathrm{~m} \text { ], }}" src="https://entranceco

We wish to see inside an atom. Assuming the atom to have a diameter of 100 pm \mathrm{\text { [1 picometer }(\mathrm{pm}), 10^{-12} \mathrm{~m} \text { ], }} this means that one must be able to resolve width of, say 10 pm. If an electron microscope is used, the minimum electron energy required is about

Option: 1

1.5 keV


Option: 2

15 KeV


Option: 3

150 keV


Option: 4

1.5 MeV


Answers (1)

best_answer

\mathrm{\begin{aligned} & \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}} \\ & \lambda^2=\frac{\mathrm{h}^2}{2 \mathrm{mK}} \\ & \mathrm{K}=\frac{\mathrm{h}^2}{2 \mathrm{~m} \lambda^2}=\frac{\left(6.6 \times 10^{-34}\right)^2}{2 \times 9.1 \times 10^{-31} \times\left(10 \times 10^{-12}\right)^2} \mathrm{~J}=\frac{6.6 \times 6.6 \times 10^{-15}}{2 \times 9.1 \times 1.6 \times 10^{-16}} \mathrm{keV}=15.1 \mathrm{KeV} \end{aligned}}

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