Answer:
The maximum wavelength of light that could liberate electrons from the aluminum metal is 303.7 nm
Explanation:
Given;
wavelength of the UV light, λ = 248 nm = 248 x 10⁻⁹ m
maximum kinetic energy of the ejected electron, K.E = 0.92 eV
let the work function of the aluminum metal = Ф
Apply photoelectric equation:
E = K.E + Ф
Where;
Ф is the minimum energy needed to eject electron the aluminum metal
E is the energy of the incident light
The energy of the incident light is calculated as follows;
The work function of the aluminum metal is calculated as;
Ф = E - K.E
Ф = 8.02 x 10⁻¹⁹ - (0.92 x 1.602 x 10⁻¹⁹)
Ф = 8.02 x 10⁻¹⁹ J - 1.474 x 10⁻¹⁹ J
Ф = 6.546 x 10⁻¹⁹ J
The maximum wavelength of light that could liberate electrons from the aluminum metal is calculated as;
Answer:
a) 316880878140.2895 Joules
b) 1056.26 Joules
c) 372.80103310622 km²
Explanation:
Energy consumption
Average rate of electrical energy consumption is 316880878140.2895 Joules
Total population = 300 million
Each person consumes 1056.26 Joules.
Efficiency = 0.85
Power from Sun
Area
Converting to km²
Area required to collect the electrical energy used in the United States is 372.80103310622 km²
The most important mathematical relationship between voltage, current and resistance in electricity is the Ohm's law: I = V/R<span>,</span>
<span>(current=voltage/resistance). We can see, that
The magnitude of the electric current is directly proportional to the voltage of the electric field.</span>
Answer:
r = 0.37 m
Explanation:
Use Coulomb's law to solve for r:
F = kq1q2/r^2
---> r^2 = kq1q2/F
= (8.99×10^9)(5.6×10^-4)(-2.1×10^-4)/(-7.7×10^3)
= 0.1373 m^2
or r = 0.37 m
Hooke’s Law, F = k e
Force = 300 x 0.153 = ..... N
