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Physics 21 Online
OpenStudy (anonymous):

A particular form of electromagnetic radiation has a frequency of 5.42x10^15 Hz. What is the wavelength is nanometers? meters?

OpenStudy (anonymous):

We are dealing with electromagnetic (EM) radiation. Therefore, we know that in a vacuum, this EM wave (like any other EM wave) propagates a speed of 3 x 10^8 m/s. This is a property of EM radiation. \[v = c = 3\times10^{8} m/s\] We are given the frequency of this EM wave: \[f = 5.42\times10^{15} s ^{-1}\] Applying the formula \[v = f \lambda\] we get \[\lambda = v / f = c/f\]\[\lambda = (3 \times 10^{8} m/s)/(5.42 \times 10^{15} s^{-1}) = 5.54 \times 10^{-8}m\] Answer: 5.54 x 10^-8 m OR 55.4 nm

OpenStudy (anonymous):

okay, so how do i know figure out: what is the energy in joules of one quantum of this radiation?

OpenStudy (anonymous):

Energy is a different matter altogether. In order to find the energy of a quantum of EM radiation (i.e. a photon), you need to use another formula: \[E = hf\] We are given that \[f = 5.42\times10^{15}s ^{-1}\]and \[h = 6.63\times10^{-34} Js\](this is Planck's constant). Therefore, \[E = (6.63\times10^{-34} Js)(5.42\times10^{15} s ^{-1}) = 3.59\times10^{-18} J\] Answer: 3.59 x 10^-18 J

OpenStudy (anonymous):

Similarly, you could use the wavelength of the EM radiation since \[c = f \lambda\]and so\[E = hf = hc/\lambda\]This would give you the same answer.

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