Question

Suppose a cube of aluminum which is 1.00 cm on a side accumulates a net charge of +2.00 pC.

(a) What percentage of the electrons originally in the cube was removed? ............................%

(b) By what percentage has the mass of the cube decreased because of this removal?............................ %

Answer 1

Net positive charge on Al cube = 2.00 X 10^(-12) Coloumbs

So total charge of electrons removed = - 2.00 X 10^(-12) coloumbs

Charge of each electron = - 1.6 X 10^(-19) Coloumbs

Total number of electrons removed = - 2.00 X 10^(-12) coloumbs/ [-1.6 X 10^(-19) Coloumbs]

= 1.25 X 10^7 electrons

Volume of cube = [1.00 X 10^(-2) m] ^3 = 10^(-6) m^3

number_density of Al = density/atomic_mass = density/26.98 gms/mole =2700Kg/m^3/ 26.98 X 10^(-3)Kg/Mole

= 100.1 X 10^3 Moles/m^3

Number density i = 100.1 X10^3moles/M^3 X 6.02 x 10^23 atoms/mole = 602.6 X 10^(26) atoms

Number density of electrons = 602.6 X 10^26 X Z =606.2 X 10^26 X 13 = 7880.6 X 10^26 /m^3

Total no.of electrons = Number density of electrons X volume of cube = 7880.6 X 10^26/m^3 X 10^(-6) m^3

= 7880.6 X 10^20 electrons

Percentage of electrons removed = (Total number removed/ Total electrons present) X 100 = =1.25 X 10^7electrons/7880.6 X 10^20 electrons =1.59 X 10^(-17)X 100

= 1.59 X 10^(-15) %

Where = No.of electrons removed

   = No of electrons in the cube initially

Q = Net charge of electrons removed ; e = charge of single electron

= 13 electrons/atom for Al

= density of Aluminium = 2.70 gms/cm^3

Vcube = 1 cm^3

= Avagadro Number

= atomic mass =26.98 gms/m

2.00 X 10^(-12)CX 26.98 gms/mole/[(13 electrons/atom)(2.70gms/cm^3)(1cm^3)(1.602X10^(-19) C )(6.02 X 10^23)]

= 1.59 X 10^(-15)%

(b)

= 2.5 X 10^(-12 ) CX 9.109 X 10^(-28)g/[1.602 X 10^(-19) C X2.70gms/cm^3X1cm^3)

=4.21 X 10^(-19) %