SATHEE: Optics Question 110

Optics Question 110

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 resolved a width of say 10 p.m. If an electron microscope is used, the minimum electron energy required is about

[AIIMS 2004]

Options:

A) 1.5 KeV

B) 15 KeV

C) 150 KeV

D) 1.5 KeV

Show Answer

Answer:

Correct Answer: B

Solution:

Wave length of the electron wave be $10 \times 10^{- 12} m$ , Using $λ = \frac{h}{\sqrt{2 m E}}$

$\Rightarrow E = \frac{h^{2}}{λ^{2} \times 2 m}$

$= \frac{{(6.63 \times 10^{- 34})}^{2}}{{(10 \times 10^{- 12})}^{2} \times 2 \times 9.1 \times 10^{- 31}} J o u l e$

$= \frac{{(6.63 \times 10^{- 34})}^{2}}{{(10 \times 10^{- 12})}^{2} \times 2 \times 9.1 \times 10^{- 31} \times 1.6 \times 10^{- 19}} e V$ = 15.1 KeV.

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Optics Question 110

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 resolved a width of say 10 p.m. If an electron microscope is used, the minimum electron energy required is about

Options:

Answer:

Solution:

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