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AP Chemistry 3.13 in-class worksheet

AP Chemistry 3.13 in-class worksheet answers
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AP Chemistry

81 Documents
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Academic year: 2022/2023
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3.

4

YOU DO:

  1. The photoelectron spectrum for Potassium is provided below.

Binding Energy per electron (x10-18 J) chemmybear/aptipin2015/PES%20Presentation%204-18-15.pdf a) Write the electron configuration for potassium.

b) Identify the valence electron(s) on the graph above. (Circle)

c) Calculate the frequency of light required to remove the valence electron(s).

  1. The energy required to eject an electron from sodium metal using the photoelectric effect is 275 kJ/mol. What is the maximum wavelength in nm needed for this to occur?

  2. Calculate the frequency of red light with a wavelength of 715 nm.

  3. The ionization energy of silver is 731 kJ/mol. Is light with a wavelength of 415 nm sufficient to remove an electron from a silver atom in the gaseous phase?

1s2 2s2 2p6 3s2 3p6 4s

v = E/h = (010^-18) / (610^-34) = 6^14 s^-

275 kJ = 275 000 J 275 000 / (610^-23) = 410^23 J/electron ^ = hc/E = (610^-34 * 310^8 * 1x10^9) / (4^-19) = 435 (nm)

c = ^v => v = c/^ = 310^810^9/715 = 4*10^14 s^

^ = hc/E = (6310^8)/731000 = 2719 => not sufficient

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AP Chemistry 3.13 in-class worksheet

Course: AP Chemistry

81 Documents
Students shared 81 documents in this course

University: Studocu University - USA

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3.12
4
YOU DO:
1) The photoelectron spectrum for Potassium is provided below.
Binding Energy per electron (x10-18 J)
http://www.chemmybear.com/aptipin2015/PES%20Presentation%204-18-15.pdf
a) Write the electron configuration for potassium.
b) Identify the valence electron(s) on the graph above. (Circle)
c) Calculate the frequency of light required to remove the valence electron(s).
2) The energy required to eject an electron from sodium metal using the photoelectric effect is 275 kJ/mol.
What is the maximum wavelength in nm needed for this to occur?
3) Calculate the frequency of red light with a wavelength of 715 nm.
4) The ionization energy of silver is 731 kJ/mol. Is light with a wavelength of 415 nm sufficient to remove an
electron from a silver atom in the gaseous phase?
1s2 2s2 2p6 3s2 3p6 4s1
v = E/h = (0.42*10^-18) / (6.626*10^-34) = 6.3x10^14 s^-1
275 kJ = 275 000 J
275 000 / (6.022*10^-23) = 4.57*10^23 J/electron
^ = hc/E = (6.626*10^-34 * 3.00*10^8 * 1x10^9) / (4.57x10^-19) = 435 (nm)
c = ^v => v = c/^ = 3*10^8*10^9/715 = 4.2*10^14 s^1
^ = hc/E = (6.626*3.00*10^8)/731000 = 2719
=> not sufficient

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