Answer :
Answer:
[tex]E_B=7x10^{-19} J[/tex]
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:
[tex]E=mc^2[/tex]
Also Planck's find and equation for the energy of a photon, given by:
[tex]E=h\nu[/tex]
[tex]h=6.63x10^{-34}Js[/tex] 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 [tex]\lambda f\tex].
[tex]E_A=2.1x10^{-18}J[/tex] (given from the problem).
2) Formulas to apply
The following relationship is important:
[tex]c=\lambda f[/tex] (1)
Where [tex]c=3x10^8 \frac{m}{s}[/tex] is the speed of the light
Solving f from equation (1) we got
[tex]f=\frac{c}{\lambda}[/tex] (2)
3) Apply the formulas
We can find the energy for each source on the following way:
[tex]E_A=hf_A=h\frac{c}{\lambda_A}[/tex]
Replacing the value given, we got:
[tex]2.1x10^{-18}J=hf_A=h\frac{c}{\lambda_A}[/tex] (3)
After this solving for [tex]/lambda_A[/tex] from equation (3) we got:
[tex]\lambda_A=\frac{hc}{2.1x10^{-18}J} [/tex] (4)
Similarly for the energy source B we have:
[tex]E_B=hf_B=h\frac{c}{\lambda_B}[/tex] (5)
For the next step we can apply the condition given, and we got: [tex]\lambda_B=3\lambda_A[/tex] (6)
Replacing the condition on equation (6) into equation (5):
[tex]E_B=hf_B=h\frac{c}{3\lambda_A}[/tex] (7)
And now, we can replace equation (4) into equation (7):
[tex]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[/tex] (7)
So the final answer would be [tex]E_B=7x10^{-19}J[/tex].