Question

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(a) 1.5 keV(b) 50 keV(c) 150 keV(d) 1.5 MeV

Answer 1

Answer:

15 keV

Explanation:

The resolving power of the electron microscope = 10pm = 10 x10-12m

That is equal to the de Broglie wavelength of the electron -

λ = 10 x 10-12m = 10-11 m.

 λ = h/p,

where h is the Planck's constant and p the momentum of the electron.

p = h/λ

= (6.62606957 × 10-34 m2 kg / s)/10-11 m.

=  6.62606957 × 10-23 m kg / s

Kinetic energy of the electron - E = p2/(2m)

where m = 9.10938291 × 10-31 kg

= (6.62606957 × 10-23 m kg / s)2/(2x9.10938291 × 10-31 kg)

= 2.4098 x 10-15J

= 2.4098 x 10-15 x 6.24150934 × 1018 eV

= 15.04 x 103 eV

= 15 keV

Answer 2

Answer:

15 ke v

Explanation:

i hope it will help you