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malfutka [58]
1 year ago
6

A 100 W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorb

s all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. At what rate are the photons from the lamp associated with the sodium light
Physics
1 answer:
ICE Princess25 [194]1 year ago
7 0

The rate at which photons from the lamp are associated with the sodium light is 2.96×10^{20}.

Power of the sodium lamp, P=100 W

The wavelength of the emitted sodium light, λ=589 nm = 589×10^{-9}m

Planck's constant, h=6.626×10^{-34}Js

Speed of light, c= 3×10^{8}m/s

(a)The energy per photon associated with the sodium light is given as

E =hcλ

E =6.626×10^{-34}×3×108/589×10^{-9}=3.37×10^{-19}J

E =\frac{3.37*10^{-19} }{1.6*10^{-19} }=2.11eV

(b)Number of photons delivered to the sphere = n

The equation for power can be written as P=nE

∴n =PE

=1003.37×10^{-19} =2.96×10^{20}photons/s

Therefore, every second, 2.96×10^{20} photons are delivered to the sphere.

To know more about photons refer to: brainly.com/question/26412768

#SPJ4

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Answer:

2354.4 Pa

40221 Pa

Explanation:

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The pressure difference would be

\Delta P=\rho gh\\\Rightarrow \Delta P=1000\times 0.24\times 9.81\\\Rightarrow \Delta P=2354.4\ Pa

The pressure difference in the first case is 2354.4 Pa

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The pressure difference in the second case is 40221 Pa

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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
-BARSIC- [3]

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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swat32

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Explanation:

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