First slide

Matter waves (DC Broglie waves)

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

Moderate

A

15 keV
B

1.5 keV
C

150 keV
D

1.5 MeV

Solution

From de-Broglie equation, we have
$λ = \frac{h}{p} = \frac{h}{mv}$
where $λ$ is wavelength, ft is Planck's constant, z is mass, p is momentum and v is velocity.
$\begin{array}{l} Given, λ = 10 pm = 10^{- 11} m, m = 9 .1 \times 10^{- 31} kg, h = 6 .6 \times 10^{- 34} Js \\ ∴ v = \frac{h}{mλ} = \frac{6 .6 \times 10^{- 34}}{9 .1 \times 10^{- 31} \times 10^{- 11}} = 7 .25 \times 10^{7} m / s \\ Energy of electron = \frac{1}{2} {mv}^{2} = \frac{1}{2} \times \frac{9 .1 \times 10^{- 31} \times {(7 .25 \times 10^{7})}^{2}}{1 .6 \times 10^{- 19}} = 15 keV \end{array}$

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