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# chapter2 - Physics 107 c f 3.10 c 8 Problem 2.5 h E 6.63.10...

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Physics 107 Problem 2.5 O. A. Pringle c3 1 0 8 . h 6.63 10 34 . λ 700 10 9 . f c λ Eh f . E 2.841 10 19 . = Joules Note I had to set the zero tolerance here. e 1.6 10 19 . eV <-> joules conversion factor E eV E e E eV 1.776 = eV ========================================================================== Physics 107 Problem 2.6 O. A. Pringle f . h 6.63 10 34 . 1 0 8 . f E h E eV 100 10 6 . eV e 1.6 10 19 . EE eV e . f E h f 2.413 10 22 . = Hz cf λ . so λ c f λ c f λ 1.243 10 14 . = meters in nanometers, λ nm λ 10 9 . λ nm 1.243 10 5 . = ========================================================================== Physics 107 Problem 2.7 O. A. Pringle h 6.63 10 34 . f 880 10 3 . photon energy f . Power is energy per time. Photons per second is energy per time divided by energy per photon. P1 1 0 3 . n P E n 1.714 10 30 . = photons per second Note that as long as I keep everything in mks units, the units automatically work out OK. ========================================================================== 1

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Physics 107 Problem 2.8 O. A. Pringle h 6.63 10 34 . E1 0 18 λ 600 10 9 . c3 1 0 8 . E photon hc . λ E photon 3.315 10 19 . = N E E photon N 3.017 = The eye can detect 3 of these photons. Note again that units work out OK if we stick with the mks system. ========================================================================== Physics 107 Problem 2.9 O. A. Pringle (a) How many photons fall per second on each square meter of the earth's surface directly facing the sun? h 6.63 10 34 . f photon 510 14 . on each square meter: P 1.4 10 3 . as in problem 2.5, number per second is power divided by energy n P hf photon . n 4.22 10 21 . = photons/s (b) What is the power output of the sun, and how many photons per second does it emit? The power per area at a radius of 1.5*10^11 m is given as 1.4x10^3 W/m^2. To answer the first part, just multiply the power per area by the area of a sphere of the given radius. r 1.5 10 11 . A4 π . r 2 . P sun PA . P sun 3.96 10 26 . = watts N P sun photon . N 1.19 10 45 . = photons per second emitted by the sun (c) How many photons per cubic meter are there near the earth? 1 0 8 . will need this in a minute This is mainly a unit conversion problem. Look at the units. number volume number s s volume . = number volume number s s m . 1 m 2 . = density number s 1 speed . 1 area . = ρ n 1 c . 1 1 2 . ρ 1.41 10 13 . = photons per cubic meter ========================================================================== 2
Physics 107 Problem 2.10 O. A. Pringle t2 0 1 0 3 . P 0.5 λ 632 10 9 . h 6.63 10 34 . c3 1 0 8 . The energy of a pulse is the power times the time. E pulse Pt . The number of photons in the pulse is the energy of the pulse divided by the energy of a photon. N E pulse hc . λ N 3.18 10 16 . = ========================================================================== Physics 107 Problem 2.11 O. A. Pringle The equation to use is K max hf . 0 .

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## This note was uploaded on 02/16/2010 for the course CHEM 550 taught by Professor Abra during the Fall '09 term at Penn State.

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chapter2 - Physics 107 c f 3.10 c 8 Problem 2.5 h E 6.63.10...

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