Question #4 4) Gold has one conduction electron per atom. its density is 19.3gr/
ID: 2009299 • Letter: Q
Question
Question #4
4) Gold has one conduction electron per atom. its density is 19.3gr/cm^3, molar mass is 197gr/mole. adn its resitivity is 2.4
2.4 x 10-8 m. A gold wire legth of 155cm and 0.247mm diameter has a potential differance of 3 volts applied accorss its endpoints . Determine the follwing:
a) The Electric Field E in the wire:
b) The number desity n:
c) the drift speed d of the electron in the wire
d) the electron collisoin frequency ƒ (ƒ =1/ )
Question #5
5. a) A rectangular block of pure sliver measures 4cm x 4cm x 20cm. determine the resistance between the opposite sides that are farthest apart:
b) Determine the dimenision (in cm) of a rectangular alumnium block (40cm long also), that would have
the same resistance between the farthest opposing sides as the silver block (part a))
Question #7
b) for different wires (materials 1 & 2) having the sames diameter d and same resistance R, show the
Explanation / Answer
Given that resisitivity of gold is = 2.4*10^-8m Length of the gold wire is L = 155cm diameter of the wire is d = 0.247 mm Area of the crosssectional wire is A = r^2 Resistance of the wire is R = L /A = 2.4*10^-8m * 155*10^-2m / (0.247 *10^-3m/2)^2 = 0.19 Potential difference is V = 3.0V a ) Electric field in the wire is E = V / d = 3.00V / 155*10^-2 m = 1.93 V / m b ) Number density n is Given that density of gold wire is = 19.3 g / cm^3 Volume of the wire is A * LVolume of wire = 0.297 * 10-6 m3 = 0.297 * 106 * 10-6cm3 = 0.297 cm3 Mass of gold wire = density * volume = 19.3 g/cm3* 0.297 cm3 = 5.732 g number of electrons is 6.023 *10^23 *5.732/197 = 0.1752*10^23 electrons number density n = ( 0.1752 * 1023 / 0.297cm3 ) = 0.590 * 1023 cm-3 c ) drift speed v_d = I / n A Q = 3.0V / 0.19 / 0.590 * 1023 cm-3 * (0.247 *10^-3m/2)^2 * 1.6*10^-19C
solve the above d ) collision frequency i = q / t time t = 0.10*10^-19 s frequency f = 1/ t = 9.8*10^19 Hz