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  • We wish to see inside an atom. Asuming the atom to have a diameter of  [1 picometer <

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

Option: 1

1.5 \mathrm{~keV}


Option: 2

15 \mathrm{~keV}


Option: 3

150 \mathrm{~keV}


Option: 4

1.5 \mathrm{~MeV}


Answers (1)

best_answer

\begin{aligned} & \text { As, } \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}} \\ & \text { or } \lambda^2=\frac{\mathrm{h}^2}{2 \mathrm{mK}} \\ & \therefore \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}=15.1 \mathrm{keV} \end{aligned}

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Ritika Harsh

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