Answer: a) 139.4 μV; b) 129.6 μV
Explanation: In order to solve this problem we have to use the Ohm law given by:
V=R*I whre R= ρ *L/A where ρ;L and A are the resistivity, length and cross section of teh wire.
Then we have:
for cooper R=1.71 *10^-8* 1.8/(0.001628)^2= 11.61 * 10^-3Ω
and for silver R= 1.58 *10^-8* 1.8/(0.001628)^2=10.80 * 10^-3Ω
Finalle we calculate the potential difference (V) for both wires:
Vcooper=11.62* 10^-3* 12 * 10^-3=139.410^-6 V
V silver= 10.80 10^-3* 12 * 10^-3=129.6 10^-6 V
Answer:
Step one : read the directions
Step two: complete the assignment
Explanation:
Answer:
D Electromagnetic and gravitational
Answer:
- The energy that must be added to the electron to move it to the third excited state is -1.153 eV
- The energy that must be added to the electron to move it to the fourth excited state is -1.181 eV
Explanation:
Given;
Energy of electron in ground state (n = 1 ) = 1.23 eV
E₁ = 1.23 eV
Eₙ = E₁ /n²
where;
E₁ is the energy of the electron in ground state
n is the energy level,
For third excited state, n = 4
E₄ = E₁ /4²
E₄ = (1.23 eV) / 16
E₄ = 0.077 eV
Change in energy level, = E₄ - E₁ = 0.077 eV - 1.23 eV = -1.153 eV
The energy that must be added to the electron to move it to the third excited state is -1.153 eV
For fourth excited state, n = 5
E₅ = E₁ /5²
E₄ = (1.23 eV) / 25
E₄ = 0.049 eV
Change in energy level, = E₅ - E₁ = 0.049 eV - 1.23 eV = -1.181 eV
The energy that must be added to the electron to move it to the fourth excited state is -1.181 eV