Answer:
The wavelength of the emitted photon is 413.6 nm
Explanation:
Given;
energy of the emitted photon, E = 3 eV = 3 x 1.602 x 10⁻¹⁹ J
speed of light, c = 3 x 10⁸ m/s
The energy of the emitted photon is given by;
E = hf
where;
h is Planck's constant = 6.626 x 10⁻³⁴ J/s
f is the frequency of the light = c / λ
λ is the wavelength
[tex]E = \frac{hc}{\lambda}\\\\\lambda = \frac{hc}{E}\\\\\lambda = \frac{(6.626*10^{-34})(3*10^8)}{3*1.602*10^{-19}}\\\\\lambda = 4.136 *10^{-7} \ m\\\\\lambda = 413.6 *10^{-9} \ m\\\\\lambda = 413.6 \ nm[/tex]
Therefore, the wavelength of the emitted photon is 413.6 nm
