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

Options

A. `15 keV`
B. `1.5 keV`
C. `150 keV`
D. `1.5 MeV`

Answer

A

Explanation

From de - Broglie equation , we have
`lambda = (h)/(p) = (h)/(mv)`
where `lambda` is wavelength , `h` is Planck's constant , `m` is mass ,`p` is momentum and `v` is velocity.
Given , `lambda = 10 pm = 10^(-11) m , m = 9.1 xx 10^(-31) kg , h = 6.6 xx 10^(-34) Js`
`:. v = (h)/(m lambda) = (6.6 xx 10^(-34))/(9.1 xx 10^(-31) xx 10^(-11))`
` = 7.25 xx 10^(7) m//s`
Energy of electron
`= (1)/(2) mv^(2) = (1)/(2) xx (9.1 xx 10^(-31) xx (7.25 xx 10^(7))^(2))/(1.6 xx 10^(-19)) = 15 keV`