hv=1E+ 1/2mu2 where v is the frequency of the UV light,and m and u are the mass and velocity of the electron, respectively. In one experiment the kinetic energy of the ejected electron from potassium is found to be 5.34 x 10-19 ] using a UV source of wavelength 162 nm. Calculate the ionization energy of potassium. How can you be sure that this ionization energy corresponds to the electron in the valence shell (that is, the most loosely held electron)?

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Chapter7: Quantum Theory Of The Atom
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Problem 7.120QP: Light with a wavelength of 405 nm fell on a strontium surface, and electrons were ejected. If the...
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hv=1E+ 1/2mu2 where v is the frequency of the UV light,and m and u are the mass and velocity of the electron, respectively. In one experiment the kinetic energy of the ejected electron from potassium is found to be 5.34 x 10-19 ] using a UV source of wavelength 162 nm. Calculate the ionization energy of potassium. How can you be sure that this ionization energy corresponds to the electron in the valence shell (that is, the most loosely held electron)?
hv = IE + 1/2mu2
where v is the frequency of the UV light, and m and u
are the mass and velocity of the electron,
respectively. In one experiment the kinetic energy of
the ejected electron from potassium is found to be
5.34 x 10-19 J using a UV source of wavelength 162
nm. Calculate the ionization energy of potassium.
How can you be sure that this ionization energy
corresponds to the electron in the valence shell (that
is, the most loosely held electron)?
Transcribed Image Text:hv = IE + 1/2mu2 where v is the frequency of the UV light, and m and u are the mass and velocity of the electron, respectively. In one experiment the kinetic energy of the ejected electron from potassium is found to be 5.34 x 10-19 J using a UV source of wavelength 162 nm. Calculate the ionization energy of potassium. How can you be sure that this ionization energy corresponds to the electron in the valence shell (that is, the most loosely held electron)?
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