Question

The most prominent line in the spectrum of aluminum is emitted at 396.15 nm. What is the frequency of this line? How much energy is in one photon of this light?

Answer 1

Well, `c=nuxxlambda`, and we get approx. `nu=8xx10^12*Hz`.....and then we must use `"the Planck relationship"`.

Explanation:

Well, `c=nuxxlambda`, where `"c=speed of light"=3.00xx10^10*cm*s^-1`.

And `lambda=396.15xx10^-9*m`.......

........`lambda=396.15xx10^-9*mxx10^2*cm*m^-1=3.96xx10^-5*cm`

And so `nu=c/lambda=(3.00xx10^10*cm*s^-1)/(3.96xx10^-5*cm)=7.58xx10^14*s^-1` or `7.58xx10^14*Hz`........

And now we use the Planck relationship......

`epsilon=hnu=6.626xx10^-34*J*sxx7.57xx10^14*s^-1=5.02xx10^-19*J`...