# Chemistry

I have this question from an old exam for review. I tried the question but can't get the right answers.

An electron in an electronically excited hydrogen atom undergoes a transition from a 6d to a 2p orbital, resulting in the emission of a photon. The photon strikes a metal surface where it is absorbed,causing an electron to be ejected having a kinetic energy of 1.32×10-19J.

(a) What is the energy (in J) of the photon emitted by the hydrogen atom?

(b)What is the wavelength (in nm) of the photon emitted by the hydrogen atom?

(c) What is the minimum energy needed to remove an electron from the metal surface?

(d) The wavelength (in nm) of the ejected electron?

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1. To start, an electron transition from a 6d to a 2p is forbidden because it violates the selection rule of delta l = +/- 1; however, if you will provide the answers that you say you can't get, I'll give it a shot.

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2. I made a huge mistake. I guess I can't count. d electrons have l = 3 while p electrons have l = 2 so delta l is 1 and there is no violation. When I first read the problem I got it in my mind that the difference was 2. Funny thing about hour minds.

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3. I figured out question A and B, now I am stuck at question C and D. The answers for question C is 3.52x10^-19 J and D is 1.35nm

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posted by Kara
4. You need the answer to a to solve d. So I obtained 4.846E-19 J for the energy of the transition (part a).
b. That translates to 410.2 nm for the wavelength.
c. K.E. = hc/wavelength - work function.
You know K.E. is 1.32E-19J. You know hc/wavelength from part a. Solve for work function which is the minimum energy required to eject an electrons.

d. You know K.E. K.E. = 1/2 mv^2. You know m (mass electron), solve for v. Then use the De Broglie equation.
wavelength = h/mv.

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5. How did you calculate the energy for a. what formulas did you use

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posted by Joe

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