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Sagot :
ANSWER
The wavelength of the photon is 0.00669 nm
STEP-BY-STEP EXPLANATION:
Given information
The energy of the photon = 2.97 x 10^-19J
Let x represents the wavelength of the photon
The first thing to do is to establish the relationship between energy and photon
Recall that,
E = hf --------------- equation 1
Where
h = Planck's constant
f = frequency
[tex]\begin{gathered} \text{Note, c = f}\lambda \\ Isolate\text{ f in the above equation} \\ f\text{ = }\frac{c}{\lambda}\text{ --------- equation 2} \end{gathered}[/tex]The next step is to substitute the value of f into equation 1, then, we have the below equation
[tex]E\text{ = h}\frac{c}{\lambda}\text{ ------equation 3}[/tex]Recall that,
h = 6.626 x 10^-34 J
c = 3 x 10^8 m/s
To find the wavelength, we need to substitute the above data into equation 3
[tex]\begin{gathered} 2.97\cdot10^{-19}\text{ = }\frac{6.626\cdot10^{-34\text{ }}\text{ x 3 }\cdot10^8}{\lambda} \\ 2.97\cdot10^{-19}\text{ = }\frac{6.626\text{ x 3 }\cdot10^{-34\text{ + 8}}}{\lambda} \\ 2.97\cdot10^{-19}\text{ = }\frac{19.878\cdot10^{-26}}{\lambda} \\ 2.97\cdot10^{-19}\text{ = }\frac{1.9878\cdot10^{-25}}{\lambda} \\ \text{Cross multiply} \\ 1.9878\cdot10^{-25}\text{ = 2.97 }\cdot10^{-19}\text{ x }\lambda \\ \text{Isolate }\lambda \\ \lambda\text{ = }\frac{1.9878\cdot10^{-25}}{2.97\cdot10^{-19}} \\ \lambda\text{ = }\frac{1.9878}{2.97}\cdot10^{-25\text{ + 19}} \\ \lambda\text{ = 0.669 }\cdot10^{-6}m \\ \lambda\text{ = 0.00669 nm} \end{gathered}[/tex]Therefore, the wavelength of the photon is 0.00669 nm
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