Question

What is the wavelength (angstroms) of a photon that has an energy of 4.38 x 10-18 j?

Answer 1

The relationship between energy and frequency of a photon is given by
$E=hf$
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:
$f= \frac{E}{h}= \frac{4.38 \cdot 10^{-18} J}{6.6 \cdot 10^{-34}Js}=6.64 \cdot 10^{15} Hz$

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

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

Answer 2

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$*10^{8}$

E =4.38*$10^{-18}$  $\frac{m^{2}*kg }{s^{2} }$

where i replaced joules for  $\frac{m^{2}*kg }{s^{2} }$

h = 6.62607004 *$10^{-34}$$\frac{m^{2}*kg }{s }$

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

But you want the solution in angstroms, so 1  meter is $10^{10}$ angstroms

so λ  = 5.7*$10^{2}$  = 570 angstrom