Question

A stellar object is emitting radiation at 1420 nm. A detector is capturing 8 107 photons per second at this wavelength. h = 6.63 x 10-34 J-s. c = 2.998 x 108 m/s.

(a) What is the total energy of the photons detected in one second?

__________J/s

(b) What is the total energy of the photons detected in one hour?

__________J/hr

Answer 1

the energy of a photon is giving:

E=hc / λ

E =energy in Joules,

h = 6.63 x 10-34 J-s. c = 2.998 x 108 m/s.

here wave length = 1420 nm = 1.42*10^-6 m

therefore ;

E=6.63⋅10−34⋅2.998 ⋅108/     1.42*10^-6

E = 13.997 *10^-20

E = 1.3997 *10^-19 J

So that's our energy, per photon.

Here A detector is capturing 8 107 photons per second

Thus it will give us the total energy per second.

1.3997 *10^-19 J *8 .10^7 photons per second

=1.1338 *10^-11

An hour has 3600 seconds(60⋅60=3600).

So E(s)⋅3600 is what we want:

1.1338 *10^-11⋅3600

=4.08 *10^-8 Joules per hour