Answer : The photon energy of this light in units of nanometers, (nm) is, [tex]4.34\times 10^{-2}nm[/tex]
Solution :
Formula used :
[tex]E=\frac{h\times c}{\lambda}[/tex]
where,
E = energy of photon = [tex]4.58\times 10^{-19}J[/tex]
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
c = speed of light = [tex]3\times 10^8m/s[/tex]
[tex]\lambda[/tex] = wavelength = ?
Now put all the given values in the above formula, we get:
[tex]4.58\times 10^{-19}J=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{\lambda}[/tex]
[tex]\lambda=4.3017\times 10^{-7}m=434.017\times 10^{-9}m=434.017nm=4.34\times 10^{-2}nm[/tex]
conversion used : [tex](1nm=1\times 10^{-9}m)[/tex]
Therefore, the photon energy of this light in units of nanometers, (nm) is, [tex]4.34\times 10^{-2}nm[/tex]
