Question

A diode laser emits at a wavelength of 989 nm . All of its output energy is absorbed in a detector that measures a total energy of 0.54 J over a period of 32 s . How many photons per second are being emitted by the laser

Answer 1

Answer:

photons per sec:

f = 3.033 E14 s-1

Explanation:

photon energy (E) is only a function of the photon wavelength:

⇒ E = h.c / λ

∴ h ( Planck contant) = 6.626070150 E-34 J.s

∴ c ( veloc. of light) = 3.0 E8 m/s

∴ λ ( wavelength) = 989 nm ( 9.89 E-7 m )

⇒ E = ((6.626070150E-34)(3.0E8))/(9.89E-7) = 2.01 E-19 J

Planck-Einstein:

• E = h.f

∴ f: frecuency

⇒ f = E/h = 2.01 E-19 J / 6.626070150 E-34 J.s

⇒ f = 3.033 E14 s-1