I think it is equal for your question
Answer:
4.1 eV
Explanation:
Kinetic energy, K = 0.8 eV = 0.8 x 1.6 x 10^-19 J = 1.28 x 10^-19 J
wavelength, λ = 253.5 nm = 253.5 x 10^-9 m
According to the Einstein energy equation
Where, E be the energy incident, Wo is the work function and K is the kinetic energy.
h = 6.634 x 10^-34 Js
c = 3 x 10^8 m/s
So, the work function, Wo = E - K
Wo = 7.85 x 10^-19 - 1.28 x 10^-19
Wo = 6.57 x 10^-19 J
Wo = 4.1 eV
Thus, the work function of the metal is 4.1 eV.
Answer:
66.375 x 10⁻⁶ C/m
Explanation:
Using Gauss's law which states that the net electric flux (∅) through a closed surface is the ratio of the enclosed charge (Q) to the permittivity (ε₀) of the medium. This can be represented as
;
∅ = Q / ε₀ -----------------(i)
Where;
∅ = 7.5 x 10⁵ Nm²/C
ε₀ = permittivity of free space (which is air, since it is enclosed in a bag) = 8.85 x 10⁻¹² Nm²/C²
Now, let's first get the charge (Q) by substituting the values above into equation (i) as follows;
7.5 x 10⁵ = Q / (8.85 x 10⁻¹²)
Solve for Q;
Q = 7.5 x 10⁵ x 8.85 x 10⁻¹²
Q = 66.375 x 10⁻⁷ C
Now, we can find the linear charge density (L) which is the ratio of the charge(Q) to the length (l) of the rod. i.e
L = Q / l ----------------------(ii)
Where;
Q = 66.375 x 10⁻⁷ C
l = length of the rod = 10.0cm = 0.1m
Substitute these values into equation (ii) as follows;
L = 66.375 x 10⁻⁷C / 0.1m
L = 66.375 x 10⁻⁶ C/m
Therefore, the linear charge density (charge per unit length) on the rod is 66.375 x 10⁻⁶ C/m.
With the switch open, there's no current in the circuit, and therefore
no voltage drop across any of the dissipative elements (the resistor
or the battery's internal impedance). So the entire battery voltage
appears across the switch, and the voltmeter reads 12.0V .
If the atom is neutral, it has the same number of protons as electrons. If there are 5 electrons, there are also 5 protons.
