Standing at a crosswalk, you hear a frequency of 550 Hz from the siren of an approaching ambulance. After the ambulance passes,
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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:
Explanation:
Assuming this problem: "Part (a) of the figure attached shows a non-conducting rod with a uniformly distributed charge +Q. The rod forms a half circle with radius R and produces an electric field of magnitude Earc at its center of curvature P. If the arc is collapsed to a point at distance R from P (Figure b), by what factor is the magnitude of the electric field at P multiplied?"
On this case the charge density is given by this formula:
assuming a half circle
We can find the force acting on the x axis with this:
We can cnvert the integral using the symmetrical property:
And we can find the electric field like this:
And the electric field just by the charge is given by:
And if we find the ratio for the two electrical fields we got:
