We wish to see inside an atom. Assume 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 energy required should be

1.5 keV

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15 keV

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150 keV

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1.5 MeV

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Solution

The correct option is B 15 keV
De broglie wavelength is given by- $λ = \frac{h}{m v}$

$⟹ v = \frac{h}{m λ}$

$λ = 10 p m = 10^{- 11} m$

Energy required= $\frac{1}{2} m v^{2} = \frac{1}{e} \frac{1}{2} m (\frac{h}{m λ})^{2} e V$

$= 15 k e V$

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We wish to see inside an atom. Assume 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 energy required should be

The correct option is B 15 keVDe broglie wavelength is given by- λ=hmv⟹v=hmλλ=10pm=10−11mEnergy required=12mv2=1e12m(hmλ)2eV=15keV

The correct option is B 15 keV
De broglie wavelength is given by- $λ = \frac{h}{m v}$

$⟹ v = \frac{h}{m λ}$

$λ = 10 p m = 10^{- 11} m$

Energy required= $\frac{1}{2} m v^{2} = \frac{1}{e} \frac{1}{2} m (\frac{h}{m λ})^{2} e V$

$= 15 k e V$