A) A 100-watt lightbulb radiates energy at a rate of 100 \\({\ m J/s}\\) (The wa
ID: 793642 • Letter: A
Question
A) A 100-watt lightbulb radiates energy at a rate of 100 ({ m J/s}) (The watt, a unit of power, or energy over time, is defined as 1 ({ m J/s})). If all of the light emitted has a wavelength of(520{ m nm}), how many photons are emitted per second? (Assume three significant figures in this calculation.) Express your answer using three significant figures.B) How much energy (in ({ m J})) is contained in 1.00 mole of 490({ m nm}) photons? Express your answer to three significant figures and include the appropriate units. C)Part A What is the de Broglie wavelength of an electron traveling at 1.48 A) A 100-watt lightbulb radiates energy at a rate of 100 ({ m J/s}) (The watt, a unit of power, or energy over time, is defined as 1 ({ m J/s})). If all of the light emitted has a wavelength of(520{ m nm}), how many photons are emitted per second? (Assume three significant figures in this calculation.) Express your answer using three significant figures.
B) How much energy (in ({ m J})) is contained in 1.00 mole of 490({ m nm}) photons? Express your answer to three significant figures and include the appropriate units. C)Part A What is the de Broglie wavelength of an electron traveling at 1.48 A) A 100-watt lightbulb radiates energy at a rate of 100 ({ m J/s}) (The watt, a unit of power, or energy over time, is defined as 1 ({ m J/s})). If all of the light emitted has a wavelength of(520{ m nm}), how many photons are emitted per second? (Assume three significant figures in this calculation.) Express your answer using three significant figures.
B) How much energy (in ({ m J})) is contained in 1.00 mole of 490({ m nm}) photons? Express your answer to three significant figures and include the appropriate units. B) How much energy (in ({ m J})) is contained in 1.00 mole of 490({ m nm}) photons? Express your answer to three significant figures and include the appropriate units. C)Part A What is the de Broglie wavelength of an electron traveling at 1.48
Explanation / Answer
a)
energy of one photon at 520 nm
= E = hv/lambda
= (6.63 x 10^-34 J.s) (3.00 x 10^8m/s) / (5.20 x 10^-7 m)
= 1.989*10^-25 / 5.20*10^-7
= 3.825 x 10 -19 J
number of photons to reach 100 J/s
= rate = (energy/s) / (energy/photon)
= (100 J/s) / (3.825 x 10-19 J/photon)
= 2.6144 x 10^20 photons/s
b)
Energy of a photon is equal to the frequency multiplied by Planck's constant.
E = f * h
frequency is equal to the speed of light divided by the wavelength.
f = v * lambda
3x10^8 / (490*10^-9 )= 6.12*10^14 Hz.
E = hf
= (6.12*10^14) * (6.63x10^-34) = 4.058*10^-19 Joule
1 mole = (6.02*10^23 )x (4.058*10^-19) = 244291.6 Joule
c)
The deBroglie wavelength :
L = h/mv;
where m is the rest mass of the electron travelling with v speed.
L = (h/mv)sqrt(1 - (v/c)^2) = h/Mv = h/P;
where Mv = mv/sqrt(1 - (v/c)^2) = P is the relativistic momentum of the rest mass m.
To relate the deBroglie wave L to only the rest mass m is to deny the contribution of the kinetic energy to the wave, which is k = Mvc. As v = 1.48*10^5 mps, the L of the electron must include the contribution of the electron's kinetic energy to the deBroglie wave.
DB wavelength is L = (6.626*10^-34/(9.109*10^-31 * 1.48*10^5))*sqrt(1 - (1.48*10^5/299*10^6)^2)
= 4.915*10^-09 meters
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