Well at the fourth half-life it would be 6.25% so if you continue calculating the answer would be B.
Answer:
b)
Explanation:
If the charge is released at rest in an electric field, it will move along the electric field, going to regions of higher electric potential if it is a negative charge (against the field direction) and towards lower potential regions if it is positive (along the field). This means that the charge will gain kinetic energy, energy that only can come from a decrease in the electric potential energy.
For a positive charge: ΔEp = q*ΔV < 0 (as ΔV < 0)
For a negative charge: ΔEp = (-q) *ΔV < 0 (as ΔV > 0)
Answer:
the charge carriers have an energy 2.8 10⁻¹⁹ J
Explanation:
The energy in a diode is conserved so the energy supplied must be equal to the energy emitted in the form of photons.
The energy of a photon is given by the Planck expression
E = h f
the speed of light, wavelength and frequency are related
c = λ f
we substitute
E = 
a red photon has a wavelength of lam = 700 nm = 700 10⁻⁹ m
we calculate the energy
E = 6.626 10⁻³⁴ 3 10⁸/700 10⁻⁹
E = 2.8397 10⁻¹⁹J
therefore the charge carriers have an energy 2.8 10⁻¹⁹ J,
We know that the Delta E + W(Work done by non-conservative
forces) = 0 (change of energy)
In here, the non-conservative force is the friction force
where f = uN (u =kinetic friction coefficient)
W= f x d = uNd ; N=mg
Delta E = 1/2 mV^2 -1/2mVi^2
umgd + 1/2mV^2 - 1/2mVi^2 = 0 (cancel out the m term)
This will then give us:
1/2Vi^2-ugd = 1/2V^2
V^2 = Vi^2 - 2ugd
So plugging in our values, will give us:
V= Sqrt (5.6^2 -2.3^2)
=sqrt (26.07)
= 5.11 m/s
