in an electron microscope we wish to study particles of diam

in an electron microscope we wish to study particles of diameter 0.1 micrometer (about 1000 times the size of single atom)a)what should be the de broglie wavelength of the electrons?b)through what potential difference should the electrons be accelerated to have that de broglie wavelength? please explain in details reason behind every step ,thank you

Solution

A. The best wavelength, which is used to view an object, is the order of the size of that object. So de broglie wavelength

lambda = 0.1*10^-6 m = 1*10^-7 m

B.

lambda = h/mv

v = h/m*lambda = 6.626*10^-34/(9.1*10^-31*1*10^-7)

v = 7281.31 m/sec

KE = PE

0.5*mv^2 = qV

V = 0.5*mv^2/q

V = 0.5*9.1*10^-31*7281.31^2/(1.6*10^-19) = 1.50*10^-4 V