mp-chapt-2-sol

# mp-chapt-2-sol - Chapter 2 Problem Solutions 1. If Planck's...

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Inha University Department of Physics Chapter 2 Problem Solutions 1. If Planck's constant were smaller than it is, would quantum phenomena be more or less conspicuous than they are now? 3. Is it correct to say that the maximum photoelectron energy KE max is proportional to the frequency n of the incident light? If not, what would a correct statement of the relationship between KE max and n be? Sol Planck’s constant gives a measure of the energy at which quantum effects are observed. If Planck’s constant had a smaller value, while all other physical quantities, such as the speed of light, remained the same, quantum effects would be seen for phenomena that occur at higher frequencies or shorter wavelengths. That is, quantum phenomena would be less conspicuous than they are now. Sol No: the relation is given in Equation (2.8) and Equation (2.9), ), ( max o h h KE n n f n - = - = So that while KE max is a linear function of the frequency n of the incident light, KE max is not proportional to the frequency.

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Inha University Department of Physics 5. Find the energy of a 700-nm photon. 7. A 1.00-kW radio transmitter operates at a frequency of 880 kHz. How many photons per second does it emit? Sol The number of photons per unit time is the total energy per unit time(the power) divided by the energy per photon, or Sol From Equation (2.11), eV. 77 1 m 10 700 m eV 10 24 1 9 - 6 . . = × × = - E Or, in terms of joules, J 10 84 2 m 10 700 m/s) 10 s)(3.0 J 10 63 6 19 9 8 34 - - - × = × × × = . . ( E . photons/s 10 72 1 Hz) 10 s)(880 J 10 63 6 J/s 10 00 1 30 3 34 3 × = × × × = = - . . ( . n h P E P
Inha University Department of Physics 9. Light from the sun arrives at the earth, an average of 1.5 x 10 11 m away, at the rate of 1.4 x 10 3 W/m 2 of area perpendicular to the direction of the light. Assume that sunlight is monochromatic with a frequency of 5.0 x 10 14 Hz. (a) How many photons fall per second on each square meter of the earth's surface directly facing the sun? (b) What is the power output of the sun, and how many photons per second does it emit? (c) How many photons per cubic meter are there near the earth? Sol (a) The number of photons per unit time per unit are will be the energy per unit time per unit area (the power per unit area, P / A ), divided by the energy per photon, or ). . . / 2 21 14 34 - 2 3 m s photons/( 10 2 4 Hz) 10 s)(5.0 J 10 (6.63 W/m 10 4 1 × = × × × = n h A P (b) With the reasonable assumption that the sun radiates uniformly in all directions, all points at the same distance from the sun should have the same flux of energy, even if there is no surface to absorb the energy. The total power is then, , . ) . ( ) . ( ) / ( W 10 0 4 m 10 5 1 4 W/m 10 4 1 4 26 2 11 2 3 2 × = × × = - p p S E R A P where R E-S is the mean Earth-Sun distance, commonly abbreviated as “1 AU,” for “astronomical unit.” The number of photons emitted per second is this power divided by the energy per photon, or . photons/s 10 2 1

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## mp-chapt-2-sol - Chapter 2 Problem Solutions 1. If Planck's...

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