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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

Naomi Aronson

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

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#........

We are now employing the Planck relationship.

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

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Answer 2

Serenity Allen

The frequency of the aluminum line at 396.15 nm is approximately (7.57 \times 10^{14}) Hz. The energy of one photon of this light is approximately (5.01 \times 10^{-19}) joules.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.