The brightest emission line in the line spectrum of potassium is at 535 nm. what is the energy of the photon emitted?

We can use the plack equation for this, E = hv, where E is energy, h is the plack constant (6.626*10⁻³⁴ J*s) and v is the frequency. We were given a wavelength, which we'll call λ, where we can use v = c/λ (c is the speed of light = 3*10⁸ m/s) to plug into the plack equation. Doing so, we get E = hc/λ, where we can plug in numbers (remember 1 nm = 10⁻⁹ m) so E = (6.626*10⁻³⁴ J*s)*(3*10⁸ m/s)/(535 *10⁻⁹ m) = 3.72 *10⁻¹⁹ J and that's the energy of our photon emission

It is the areas with lines. this may sound a little confusing, but i hope it !