Question

a. A sculptor has asked you to help electroplate gold onto a brass statue. You know that the charge carriers in the ionic solution are singly ionized gold ions, and you've calculated that you must deposit 0.50 g of gold to reach the necessary thickness. How much current do you need, in mA, to plate the statue in 4.0 hours?

b. A metallic wire has a diameter of 4.12mm. When the current in the wire is 8.00A, the drift velocity is 5.40×10−5m/s. What is the density of free electrons in the metal?

c. A 1.3×10⁻³ V/m electric field creates a 4.4×10¹⁷ electrons/s current in a 1.4-mm-diameter aluminum wire. What is the drift speed? What is the mean time between collisions for electrons in this wire?

d. A 18-cm-long nichrome wire is connected across the terminals of a 1.5 V battery.What is the electric field inside the wire? What is the current density inside the wire? If the current in the wire is 2.0 A , what is the wire's diameter?

Answer 1

answer) a) first we have to find number of moles in gold atoms

number of moles=0.50g/196.9666g/mole=0.002539 mole

so the number of atoms =0.002539*6.022*10²³=1.52869*10²¹atoms

now charge =1.52869*10²¹*1.6*10^-19C=244.5897 C

now we have the formula

I=Q/t=244.5897C/4*3600 s=16.99 mA

so the answer is 17 mA or 17.0 mA

b) for density we have the formula

n=I/qvA

A=pir²

r=d/2=4.12mm/2=2.06mm=2.06*10^-3m

so now putting all the values in above eqn

n=8/1.60*10^-19*5.4*10^-5m/s*pi*(2.06*10^-3)²=6.95*10²⁸m^-3

so the answer is 6.9*10²⁸m^-3 or 7.0*10²⁸m^-3 or 6.95*10²⁸m^-3

d) here we have the formula

E=V/m=1.5/0.18m=8.3 V/m

so the electric field is 8.3 V/m or 8.33 V/m

we have the formula for current density=E=8.33V/m*6.7*10⁵ohm^-1m^-1=5.6*10⁶ A/m²

so the current density=5.6*10⁶A/m² or 5.58*10⁶A/m²

the wire diameter

we have the formula

current density=current/A

A=Current/Current density=2.0/5.58*10⁶

pir²=2/5.58*10⁶

r²=11.407997*10^-8

r=3.38*10^-4m

d=2r=6.76*10^-4m

so diameter=6.8*10^-4m or 6.76*10^-4m or 0.68 mm or 0.676mm

( for question c , i need the value for free electron density of aluminium which is missing)