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skyluke89
The relationship between energy and frequency of a photon is given by
[tex]E=hf[/tex]
where E is the energy, h is the Planck constant and f is the photon frequency. By re-arranging the equation and using the photon energy, we can calculate its frequency:
[tex]f= \frac{E}{h}= \frac{4.38 \cdot 10^{-18} J}{6.6 \cdot 10^{-34}Js}=6.64 \cdot 10^{15} Hz [/tex]

Then we know that the photon travels at speed of light, c, so we can find its wavelength by using
[tex]\lambda= \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{6.64 \cdot 10^{15}Hz}=4.52 \cdot 10^{-8} m [/tex]

And since 1 A (angstrom) corresponds to [tex]10^{-15} m[/tex], the wavelength expressed in angstroms is
[tex]\lambda= \frac{ 4.52 \cdot 10^{-8} m}{10^{-15} m/A} = 4.52 \cdot 10^7 A [/tex]
facundo3141592

Answer: The relationship between energy and frequency of a photon is given by E = h*f.

E is the energy, h is planck constant and f is the frequency and f= c/λ.

But i want wavelength, so i write this equation as E= h*c/λ.

now

c=3.8[tex]*10^{8}[/tex]

E =4.38*[tex]10^{-18} [/tex]  [tex]\frac{m^{2}*kg }{s^{2} }[/tex]

where i replaced joules for  [tex]\frac{m^{2}*kg }{s^{2} }[/tex]

h = 6.62607004 *[tex]10^{-34}[/tex][tex]\frac{m^{2}*kg }{s }[/tex]

then λ = h*c/E =  [tex]\frac{6.62607004 *10^{-34}*3.8*10^{8}m/s}{4.38*10^{-18} \frac{m^{2}*kg }{s^{2} }}[/tex] = 5.7*[tex]10^{-8}[/tex]  m

But you want the solution in angstroms, so 1  meter is [tex]10^{10}[/tex] angstroms

so λ  = 5.7*[tex]10^{2}[/tex]  = 570 angstrom

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