If light of wavelength 307 nm is incident on uranium metal, electrons are ejecte

Question

If light of wavelength 307 nm is incident on uranium metal, electrons are ejected from the metal with a velocity of 5.4 x l0^5 m/s. This phenomenon is called the photoelectric effect.

1)Calculate the incident photon frequency in Hertz (Hz).

2)Calculate the incident photon energy.

3)The de Broglie equation was key to the development of the concept of quantum mechanics of the atom by

Schrodinger. Calculate the de Brogue wavelength of the ejected electrons.

4)Calculate the binding energy of uranium, the minimum energy needed to eject electrons from uranium metal.

Express your answer in kJ/mol.

Explanation / Answer

1) Frequency = velocity of light/wavelength = (3*108)/(307*10-9) = 9.771*1014 s-1

2) Energy of the incident photon = Planck's constant*frequency = (6.63*10-34)*( 9.771*1014) = 6.48*!0-19 J

3) de broglie wavelength = Planck's Constant/(mass*velocity) =  (6.63*10-34)/{(9.1*10-31)*(5.4*105)} = 1.35*10-9 m = 1.35 nm

4) Binding energy = Total incident energy - K.E = 6.48*!0-19 J - {(1/2)*(9.1*10-31)*(5.4*105)2}

= 6.48*!0-19 J - 1.33*10-19 J = 5.15*10-19 J

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