How do you calculate the energy of a photon of radiation with a frequency of $8.5 \times 10^{14}$ $8.5 times 10^14$ $H z$ $Hz$ ?

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Answer 1 · 2017-04-09T04:37:24

Answer rounded according to 2 sig figs: $5.6 \times 10^{- 19} J$ $5.6 xx 10^-19"J"$

$E = h \times f$ $E= h xx f$

$E$ $E$ is energy in joules
$h$ $h$ is Planck's constant $= 6.626 \times 10^{- 34} \frac{m^{2} kg}{s}$ $= 6.626 xx 10^-34("m"^2"kg")/"s"$
$f$ $f$ is frequency, in $Hz$ $"Hz"$ or $s^{- 1}$ $"s"^(-1)$

So

$E = 6.626 \times 10^{- 34} \frac{m^{2} kg}{s} \times (8.5 \times 10^{14}) s^{- 1}$ $E= 6.626 xx 10^-34 ("m"^2 "kg")/"s" xx (8.5 xx 10^14)"s"^(-1)$

$E = 5.6321 \times 10^{- 19} J$ $E= 5.6321 xx 10^-19" J"$

How do you calculate the energy of a photon of radiation with a frequency of 8.5 times 10^148.5×10148.5 times 10^14 HzHzHz?