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

Answer:

$E_B=7x10^{-19} J$

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:

$E=h\nu$

$h=6.63x10^{-34}Js$ 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 $\lambda f\tex].[tex]E_A=2.1x10^{-18}J$ (given from the problem).

2) Formulas to apply

The following relationship is important:

$c=\lambda f$   (1)

Where $c=3x10^8 \frac{m}{s}$ is the speed of the light

Solving f from equation (1) we got

$f=\frac{c}{\lambda}$   (2)

3) Apply the formulas

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

$E_A=hf_A=h\frac{c}{\lambda_A}$

Replacing the value given, we got:

$2.1x10^{-18}J=hf_A=h\frac{c}{\lambda_A}$   (3)

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

$\lambda_A=\frac{hc}{2.1x10^{-18}J}$   (4)

Similarly for the energy source B we have:

$E_B=hf_B=h\frac{c}{\lambda_B}$   (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):

$E_B=hf_B=h\frac{c}{3\lambda_A}$   (7)

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

$E_B=hf_B=h\frac{c}{3\lambda_A}=\frac{hc}{3}\frac{2.1x10^{-18}J}{hc}=\frac{2.1x10^{-18 J}}{3}=7x10^{-19}J$   (7)

So the final answer would be $E_B=7x10^{-19}J$.