Question

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

Answer 1

`E_B=7x10^(-19) J`

Step-by-step explanation:

1) Key concepts and notation

For this case we can use the Einstein's formula for the energy of a photon. The original formula from Einstein's is:

`E=mc^2`

Also Planck's find and equation for the energy of a photon, given by:

`E=h\\u`

`h=6.63x10^(-34)Js` representing the Planck's constant

f= the frequency

Based on the above formula, the energy of a photon depends only on its wavelength and frequency. The energy is proportional to
`\lambda f\tex].</p><p>[tex]E_A=2.1x10^(-18)J` (given from the problem).

2) Formulas to apply

The following relationship is important:

`c=\lambda f` (1)

Where
`c=3x10^8 (m)/(s)` is the speed of the light

Solving f from equation (1) we got

`f=(c)/(\lambda)` (2)

3) Apply the formulas

We can find the energy for each source on the following way:

`E_A=hf_A=h(c)/(\lambda_A)`

Replacing the value given, we got:

`2.1x10^(-18)J=hf_A=h(c)/(\lambda_A)` (3)

After this solving for
`/lambda_A` from equation (3) we got:

`\lambda_A=(hc)/(2.1x10^(-18)J)` (4)

Similarly for the energy source B we have:

`E_B=hf_B=h(c)/(\lambda_B)` (5)

For the next step we can apply the condition given, and we got:
`\lambda_B=3\lambda_A` (6)

Replacing the condition on equation (6) into equation (5):

`E_B=hf_B=h(c)/(3\lambda_A)` (7)

And now, we can replace equation (4) into equation (7):

`E_B=hf_B=h(c)/(3\lambda_A)=(hc)/(3)(2.1x10^(-18)J)/(hc)=(2.1x10^(-18 J))/(3)=7x10^(-19)J` (7)

So the final answer would be
`E_B=7x10^(-19)J`.