Question

We wish to see inside an atom. Assuming the atom to have a diameter of 100 pm, this means that one must be able to resolve a width of say 10 pm. If an electron microscope is used, the minimum electron energy required is about –

• A. 1.5 keV
• B. 15 keV
• C. 150 keV
• D. 1.5 MeV

Answer 1

Answer: (B)

15 keV

Answer 2

λ = $\frac { 12.27 } { \sqrt { V } }$ Å

$\frac { 12.27 \times 10 ^ { - 10 } } { 10 \times 10 ^ { - 12 } }$

V = 1.5 kV

E = 1.5 keV