A 100 W incandescent light bulb converts approximately 2.5% of the electrical en

Question

A 100 W incandescent light bulb converts approximately 2.5% of the electrical energy supplied to it into visible light. Assume that the average wavelength of the emitted light is ? = 540 nm, and that the light is radiated uniformily in all directions.

1)

What is the energy of each photon in eV?

E =

eV

2)

How many photons per second, N, would enter an aperture of area A = 2cm2 located a distance D = 5 m from the light bulb?

N =

photons/s

3)

Suppose, instead, that the average photon wavelength is 810 nm. How many photons per second, N', would enter the aperture?

N' =

photons/s

Explanation / Answer

1) (1240eV*nm)/540.0nm

2) ((2.5/(4pi*r^2))*D)/(1.602E-19)*(energy of each photon from question #1)

3) Rinse. wash. repeat. Except lather it with lamba=810nm instead of lamba=540nm.

Because I love the CHIEF, and the great state of Illinois:

1) 2.296 eV

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