Given an electron beam whose electrons have kinetic energy of 5.00 kev , what is the minimum wavelength λmin of light radiated b
1 answer:
Wavelength and frequency are inversely proportional, therefore the minimum wavelength occurs when the emitted radiation has maximum frequency.
The energy of the emitted photons is
(1)
where h is the Planck constant and f is the frequency. The maximum energy that can be given to the emitted photons corresponds to the kinetic energy of the electrons (in fact, if they are completely stopped, they give all their kinetic energy to the photons). Converting the energy in Joules:
from eq.(1) we find the corresponding frequency of the photons:
This is the maximum frequency the emitted photons can have, and by using the basic relationship between frequency, wavelength and speed of light c, we can find the corresponding wavelength:
And this is the minimum wavelength of the radiation emitted by this electron beam.
The energy of the emitted photons is
where h is the Planck constant and f is the frequency. The maximum energy that can be given to the emitted photons corresponds to the kinetic energy of the electrons (in fact, if they are completely stopped, they give all their kinetic energy to the photons). Converting the energy in Joules:
from eq.(1) we find the corresponding frequency of the photons:
This is the maximum frequency the emitted photons can have, and by using the basic relationship between frequency, wavelength and speed of light c, we can find the corresponding wavelength:
And this is the minimum wavelength of the radiation emitted by this electron beam.
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Answer:
q = q₀ sin (wt)
Explanation:
In your statement it is not clear the type of circuit you are referring to, there are two possibilities.
1) The circuit of this problem is a system formed by an Ac voltage source and a capacitor, in this case all the voltage of the source is equal to the voltage at the terminals of the capacitor
ΔV = Δ
we assume that the source has a voltage of the form
ΔV = ΔV₀o sin wt
The capacitance of a capacitor is
C = q / ΔV
q = C ΔV sin wt
the current in the circuit is
i = dq / dt
i = c ΔV₀ w cos wt
if we use
cos wt = sin (wt + π / 2)
we make this change by being a resonant oscillation
we substitute
i = w C ΔV₀ sin (wt + π/2)
With this answer we see that the current in capacitor has a phase factor of π/2 with respect to the current
2) Another possible circuit is an LC circuit.
In this case the voltage alternates between the inductor and the capacitor
V_{L} + V_{C} = 0
L di / dt + q / C = 0
the current is
i = dq / dt
they ask us for a solution so that
L d²q / dt² + 1 / C q = 0
d²q / dt² + 1 / LC q = 0
this is a quadratic differential equation with solution of the form
q = A sin (wt + Ф)
to find the constant we derive the proposed solution and enter it into the equation
di / dt = Aw cos (wt + Ф)
d²i / dt²= - A w² sin (wt + Ф)
- A w² + 1 /LC A = 0
w = √ (1 / LC)
To find the phase factor, for this we use the initial conditions for t = 0
in the case of condensate for t = or the charge is zero
0 = A sin Ф
Ф = 0
q = q₀ sin (wt)