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What is the energy of a photon of infrared radiation with a frequency of 2.53 × 1012 Hz? Planck’s constant is
6.63 × 10–34 J • s.

A. 1.68 × 1023 J      B. 1.68 × 1047 J     C. 1.68 × 10–21 J      D. 1.68 × 10–45 J

Respuesta :

Answer : The energy of a photon of infrared radiation is, [tex]1.68\times 10^{-21}J[/tex]

Solution :

Formula used :

[tex]E=h\times \nu[/tex]

where,

E = energy of a photon = ?

h = Planck's constant = [tex]6.63\times 10^{-34}Js[/tex]

[tex]\nu[/tex] = frequency of a photon = [tex]2.53\times 10^{12}Hz=2.53\times 10^{12}s^{-1}[/tex]

Now put all the given values in this formula, we get the energy of a photon.

[tex]E=(6.63\times 10^{-34}Js)\times (2.53\times 10^{12}s^{-1})[/tex]

[tex]E=1.68\times 10^{-21}J[/tex]

Therefore, the energy of a photon of infrared radiation is, [tex]1.68\times 10^{-21}J[/tex]

Answer:

C on edge

Explanation:

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