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3 years ago

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The wavelengths of light emitted by a firefly span the visible spectrum but have maximum intensity near 550 nm. A typical flash

lasts for 100 ms and has a power of 1.2 mW. If we assume that all of the light is emitted at the peak-intensity wavelength of 550 nm, how many photons are emitted in one flash?

1 answer:

Reil [10]3 years ago

3 0

Answer:

$3.3\cdot 10^{15}$

Explanation:

First of all, let's calculate the energy of a single photon of wavelength

$\lambda=550 nm=5.5\cdot 10^{-7}m$

which is given by

$E_1 = \frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.5\cdot 10^{-7} m}=3.6\cdot 10^{-19} J$

The power of the flash is

$P=1.2 mW=0.0012 W$

and the time it lasts is

$t=100 ms=0.1 s$

so the total energy delivered in one flash is

$E=Pt=(0.0012 W)(0.1 s)=1.2\cdot 10^{-3}J$

This energy contains exactly N photons each of energy $E_1$ , so the number of photons emitted in one flash is

$N=\frac{E}{E_1}=\frac{1.2\cdot 10^{-3} W}{3.6\cdot 10^{-19}J}=3.3\cdot 10^{15}$

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Answer:

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Explanation:

Given:

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So, the net acceleration acting on the package will be up the incline and is equal to:

$a=\mu g\cos\theta-g\sin\theta$ ----------------- 1

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$-4.00=24.0a\\\\a=-\frac{4}{24}=-\frac{1}{6}\ m/s^2$

Now, plug in the value of 'a' in equation (1). This gives,

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Plug in all the given values and solve for $\mu$ . This gives,

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Answer:

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Answer:

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Substituting into equation 2,

e = √(2×6.48/450)

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<em>Thus the distance required = 16.97 cm</em>

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