frequency and energy associated withphoton of radiation having wavelength 6200 armstrong?

Answer 1

The frequency (f) of light with a known wavelength (lambda) is given as,
f = c / lambda
Where c is the speed of light in a vacuum, about 3 E8 m/s.

1 Angstrom = 1 E-10 meters,
6200 Angstroms = 6.2 E-7 m

So the frequency is,
f = 3 E8 m/s / 6.2 E-7 m
f = 4.84 E14 Hz


The energy contained within a single photon of a given frequency is given by Planck's equation,
E = hv
where E is the energy, h is Planck's constant (6.63 E-34 J s), and v is the frequency of the light (it is the same as f above, but this time I used a Greek letter to denote it).

We already know the frequency from above, and h is a constant, so we can just plug in.

E = 6.63 E-34 J s * 4.84 E14 s^-1
E = 3.2 E-19 Joules