Physics for Scientists and Engineers with Modern Physics, Technology Update
Physics for Scientists and Engineers with Modern Physics, Technology Update
9th Edition
ISBN: 9781305401969
Author: SERWAY, Raymond A.; Jewett, John W.
Publisher: Cengage Learning
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Chapter 40, Problem 11P
To determine

The number of photons per second escaping the opening and having wavelengths between 500 nm and 501 nm .

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Answer to Problem 11P

The number of photons per second escaping the opening and having wavelengths between 500 nm and 501 nm  is 1.30×1015 photons/s .

Explanation of Solution

The wavelength range is between 500 nm and 501 nm . Since this range of wavelength is small, the wavelength in the Plank’s law can be taken as the average of these two wavelengths.

Write the Plank’s radiation law.

  I(λ¯,T)=2πhc2λ¯5(ehc/λ¯kBT1)                                                                                          (I)

Here, I(λ¯,T) is the intensity per wavelength, h is the Planck’s constant, c is the speed of the photon, λ¯ is the average wavelength of the radiation emitted, kB is the Boltzmann constant and T is the temperature of the blackbody.

Write the equation for λ¯ .

  λ¯=λ1+λ22                                                                                                             (II)

Here, λ1 is the lower value of the wavelength and λ2 is the higher value of the wavelength.

Put equation (II) in equation (I).

  I(λ¯,T)=2πhc2[(λ1+λ2)/2]5(ehc/[(λ1+λ2)/2]kBT1)                                                           (III)

The energy per time leaving the hole can be determined by taking the product of the intensity per wavelength, area and the wavelength range.

Write the equation for the energy per time leaving the hole.

  P=I(λ¯,T)(λ2λ1)A                                                                                           (IV)

Here, P is the energy per time leaving the hole and A is the area of the hole.

Write the equation for the area of the hole.

  A=π(d2)2=πd24                                                                                                           (V)

Here, d is the diameter of the hole.

Put equations (III) and (V) in equation (IV).

  P=2πhc2[(λ1+λ2)/2]5(ehc/[(λ1+λ2)/2]kBT1)(λ2λ1)πd24=16π2hc2(λ1+λ2)5(e2hc/(λ1+λ2)kBT1)(λ2λ1)d2                                              (VI)

Write the equation for the average photon energy.

  E=hf                                                                                                                   (VII)

Here, E¯ is the average photon energy and f¯ is the average photon frequency.

Write the equation for f¯ .

  f¯=cλ¯

Put equation (II) in the above equation.

  f¯=c(λ1+λ2)/2=2cλ1+λ2

Put the above equation in equation (VII).

  E=h(2cλ1+λ2)=2hcλ1+λ2                                                                                                    (VIII)

Write the equation for the number of photons per second escaping the opening.

  n=PE¯

Here, n is the number of photons per second escaping the opening.

Put equations (VI) and (VIII) in the above equation.

  n=16π2hc2(λ1+λ2)5(e2hc/(λ1+λ2)kBT1)(λ2λ1)d22hcλ1+λ2=8π2cd2(λ2λ1)(λ1+λ2)4(e2hc/(λ1+λ2)kBT1)                                                           (IX)

Conclusion:

It is given that the temperature of the blackbody is 7500 K and the diameter of the opening is 0.0500 mm . The value of h is 6.626×1034 Js , the speed of photon is 3.00×108 m/s and the value of kB is 1.38×1023 J/K .

Substitute  , 3.00×108 m/s for c , 0.0500 mm for d , 501 nm for λ2 , 500 nm for λ1 , 6.626×1034 Js for h , 1.38×1023 J/K for kB and 7500 K for T in equation (IX) to find n .

    n=8π2(3.00×108 m/s)(0.0500 mm1 m1000 mm)2(501 nm1 m109 nm500 nm1 m109 nm)(500 nm1 m109 nm+501 nm1 m109 nm)4(e2(6.626×1034 Js)(3.00×108 m/s)/(500 nm1 m109 nm+501 nm1 m109 nm)(1.38×1023 J/K)(7500 K)1)=1.30×1015 photons/s

Therefore, the number of photons per second escaping the opening and having wavelengths between 500 nm and 501 nm  is 1.30×1015 photons/s .

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Chapter 40 Solutions

Physics for Scientists and Engineers with Modern Physics, Technology Update

Ch. 40 - Prob. 4OQCh. 40 - Prob. 5OQCh. 40 - Prob. 6OQCh. 40 - Prob. 7OQCh. 40 - Prob. 8OQCh. 40 - Prob. 9OQCh. 40 - Prob. 10OQCh. 40 - Prob. 11OQCh. 40 - Prob. 12OQCh. 40 - Prob. 13OQCh. 40 - Prob. 14OQCh. 40 - Prob. 1CQCh. 40 - Prob. 2CQCh. 40 - Prob. 3CQCh. 40 - Prob. 4CQCh. 40 - Prob. 5CQCh. 40 - Prob. 6CQCh. 40 - Prob. 7CQCh. 40 - Prob. 8CQCh. 40 - Prob. 9CQCh. 40 - Prob. 10CQCh. 40 - Prob. 11CQCh. 40 - Prob. 12CQCh. 40 - Prob. 13CQCh. 40 - Prob. 14CQCh. 40 - Prob. 15CQCh. 40 - Prob. 16CQCh. 40 - Prob. 17CQCh. 40 - The temperature of an electric heating element is...Ch. 40 - Prob. 2PCh. 40 - Prob. 3PCh. 40 - Prob. 4PCh. 40 - Prob. 5PCh. 40 - Prob. 6PCh. 40 - Prob. 7PCh. 40 - Prob. 8PCh. 40 - Prob. 9PCh. 40 - Prob. 10PCh. 40 - Prob. 11PCh. 40 - Prob. 12PCh. 40 - Prob. 14PCh. 40 - Prob. 15PCh. 40 - Prob. 16PCh. 40 - Prob. 17PCh. 40 - Prob. 18PCh. 40 - Prob. 19PCh. 40 - Prob. 20PCh. 40 - Prob. 21PCh. 40 - Prob. 22PCh. 40 - Prob. 23PCh. 40 - Prob. 25PCh. 40 - Prob. 26PCh. 40 - Prob. 27PCh. 40 - Prob. 28PCh. 40 - Prob. 29PCh. 40 - Prob. 30PCh. 40 - Prob. 31PCh. 40 - Prob. 32PCh. 40 - Prob. 33PCh. 40 - Prob. 34PCh. 40 - Prob. 36PCh. 40 - Prob. 37PCh. 40 - Prob. 38PCh. 40 - Prob. 39PCh. 40 - Prob. 40PCh. 40 - Prob. 41PCh. 40 - Prob. 42PCh. 40 - Prob. 43PCh. 40 - Prob. 45PCh. 40 - Prob. 46PCh. 40 - Prob. 47PCh. 40 - Prob. 48PCh. 40 - Prob. 49PCh. 40 - Prob. 50PCh. 40 - Prob. 51PCh. 40 - Prob. 52PCh. 40 - Prob. 53PCh. 40 - Prob. 54PCh. 40 - Prob. 55PCh. 40 - Prob. 56PCh. 40 - Prob. 57PCh. 40 - Prob. 58PCh. 40 - Prob. 59PCh. 40 - Prob. 60APCh. 40 - Prob. 61APCh. 40 - Prob. 62APCh. 40 - Prob. 63APCh. 40 - Prob. 64APCh. 40 - Prob. 65APCh. 40 - Prob. 66APCh. 40 - Prob. 67APCh. 40 - Prob. 68APCh. 40 - Prob. 69APCh. 40 - Prob. 70APCh. 40 - Prob. 71APCh. 40 - Prob. 72CPCh. 40 - Prob. 73CPCh. 40 - Prob. 74CPCh. 40 - Prob. 75CPCh. 40 - Prob. 76CP
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