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2 years ago

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The voltage across the diode indicates the energy given to charge carriers (electrons and holes, but more about that later in th

e course.) If the diode turns on at X volts, then the charge carriers are getting X electron-volts of energy. The charge carriers can give up this energy by emitting a photon. How does the energy the charge carriers get at the turn-on voltage compare to the energy of a red photon

1 answer:

Elis [28]2 years ago

3 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 = $\frac{h \ c}{\lambda}$

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,

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Answer:

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A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo

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Answer:

It would take $\tau(\ln 9 - \ln 8)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{8}{9} \, q_0$ .

It would take $\tau(\ln 9 - \ln 7)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{7}{9}\, q_0$ .

Note that $\ln 9 = 2\,\ln 3$ , and that $\ln 8 = 3\, \ln 2$ .

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is $\tau$ , and the initial charge of the capacitor be $q_0$ . Then at time $t$ , the charge stored in the capacitor would be:

$\displaystyle q(t) = q_0 \, e^{-t / \tau}$ .

<h3>a)</h3>

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{8}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9$ .

$t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8)$ .

<h3>b)</h3>

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{7}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9$ .

$t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7)$ .

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