1. A 100-watt lightbulb radiates energy at a rate of 100 J/s (The watt, a unit o
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Question
1. A 100-watt lightbulb radiates energy at a rate of 100 J/s (The watt, a unit of power, or energy over time, is defined as 1 J/s). If all of the light emitted has a wavelength of 505nm, how many photons are emitted per second? (Assume three significant figures in this calculation.)
2.Calculate the de Broglie wavelength of a 143-g baseball traveling at 80 mph .
3.An electron undergoes a transition from an initial (ni) to a final (nf ) energy state. The energies of the ni and nf energy states are 2.179×1018 J and 8.720×1020 J, respectively.
Calculate the wavelength () of the light in nanometers (nm) corresponding to the energy change (E) value of this transition. You can use the following values for your calculations:
Plancks constant (h)speed of light (c)1 m===6.626×1034 Js2.998×108 m/s109 nm
Explanation / Answer
Q. 1)
Power = 100 w (J/s)
Wavelength of a photon = 505 nm = 505 E -9 m
Calculation of energy of a photon
E = hc/l = sf
Here E is energy in J , h is planks constant = 6.626 E-34 ,
c = speed of light = 3.0 E8 m/s
l = is wavelength in m
E = 6.626 E-34 Js x 3.0 E8 m/s )/ 505 E-9 m
= 3.94 E-19 J
Calculation of number of photon
Emission of energy per s = 100 J
Number of photon in 100 J = 100 J x 1 photon / 3.94 E-19 J
= 2.54 E20 photon
Since the energy emission is for one second so photon emitted per second
= 2.54 E20 photon
Q. 2 )
l = h/P
Here l = is wavelength in m , h is planks constant
P : momentum = mv
Here m is mass in kg and v is velocity in m/s
Calculation of mass in kg
= 143 g x 1 kg / 1000 g
= 0.143 kg
V = 80 m/hr
Conversion of velocity to m/s
= (80 m/ hr )x (1hr/3600 s)
= 0.022 m/s
l = 6.626 E-34 J per s / (0.143 kg x 0.0222 m/s)
= 2.085 E-31 m
3.
l = hc / Delta E
Delt a E = E(final) – E(initial)
l = 6.626 E-34 Js x 2.998 E8 m per s )/ [ ( -8.720 E-20 J ) – (- 2.179 E-18 ) ]
= 9.50 E-8 m
l = 9.50 E-8 m
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