Question

Constants Periodic Table A scientist wants to use an electron microscope to observe details on the order of 0.80 nm Part A Through what potential difference must the electrons be accelerated from rest so that they have a de Broglie wavelength of this magnitude? Express your answer using two significant figures. V AE O ? V = V Submit Request Answer Provide Feedback Next >

Answer 1

Solution:

Given that

The deBrogle's wavelength,

X=0.80 nm

We have to find the potential difference,

=?

we know the equation for deBrogle's wavelength

$\lambda =\frac{h}{mv}$

Where

$h=\mathrm{Plank's \; constant}=6.625\times 10^{-34}\; \mathrm{J.s}$

m = mass of electron = 9.1 x 10 31 kg

$v=\mathrm{velocity\; of \; electron}$

Now, susbtitute all the known values

$v =\frac{h}{m\lambda }$

$v =\frac{6.625\times 10^{-34}}{9.1\times 10^{-31}\times 0.80\times 10^{-9} }$

$v =9.1\times 10^{5}\; \mathrm{m/s}$

Now, calculate the kintentic energy these electrons

$E=\frac{1}{2}mv^{2}$

$E=\frac{1}{2}(9.1\times 10^{-31})(9.1\times 10^{5})^{2}$

$E=3.7679\times 10^{-19}\; \mathrm{J}$

Now, potential difference cab be calculated as follows

$\frac{1}{2}mv^{2}=eV$

where

$e=\mathrm{charge \; of \; electron}=1.6\times 10^{-19}\; \mathrm{C}$

$V=\mathrm{potential \; difference}$

Substitute the known values

$eV=3.7679\times 10^{-19}$

$V=\frac{3.7679\times 10^{-19}}{e}$

$V=\frac{3.7679\times 10^{-19}}{1.6\times 10^{19}}$