Question

Two sources produce electromagnetic waves. Source A produces a wavelength that is three times the wavelength produced by source B. Each photon from source B has an energy of 2.1*10^-18 J. What is the energy of a photon from source A?

Answer 1

I found `7xx10^-19J`

Explanation:

We can use Einstein's formula for the energy of a photon:
`E=h*nu`
Where:
`h=6.63xx10^-34Js` is Planck's constant:
`nu=`frequency.
Also we remember the relationship between wavelngth `lambda` and frequenct `nu` through the speed of light in vacuum `c=3xx10^8m/s` as:
`c=lambda*nu`

So the energy from source A will be:
`E_A=h*nu_A=hc/lambda_A`
and for source B:
`E_B=2,1xx10^-18=h*nu_B=hc/lambda_B`
so: `lambda_B=(hc)/(2.1xx10^-18)`

BUT:
`lambda_A=3lambda_B`
so:
`E_A=h*nu_A=hc/lambda_A=(hc)/(3lambda_B)=(hc)/3*(2.1xx10^-18)/(hc)=(2.1xx10^-18)/3=7xx10^-19J`