To find the energy of a photon of infrared radiation with a given frequency, we use the equation derived from Planck’s formula for the energy of a photon:
[tex]\[ E = h \cdot f \][/tex]
where
- [tex]\( E \)[/tex] is the energy of the photon,
- [tex]\( h \)[/tex] is Planck's constant, which is [tex]\( 6.63 \times 10^{-34} \, J \cdot s \)[/tex],
- [tex]\( f \)[/tex] is the frequency of the radiation (given as [tex]\( 2.53 \times 10^{12} \, Hz \)[/tex]).
Plugging in the values:
[tex]\[ E = (6.63 \times 10^{-34} \, J \cdot s) \times (2.53 \times 10^{12} \, Hz) \][/tex]
Performing the multiplication:
[tex]\[ E = 6.63 \times 2.53 \times 10^{-34 + 12} \, J \][/tex]
[tex]\[ E = 16.7739 \times 10^{-22} \, J \][/tex]
[tex]\[ E = 1.67739 \times 10^{-21} \, J \][/tex]
We see that [tex]\( 1.67739 \times 10^{-21} \, J \)[/tex] is very close to [tex]\( 1.68 \times 10^{-21} \, J \)[/tex].
Therefore, the correct answer is:
[tex]\[ \boxed{1.68 \times 10^{-21} \, J} \][/tex]
