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Radiation of frequency 9.00 × 10^15 Hz is incident on a clean metal surface - Scottish Highers Physics - Question 12 - 2016

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Radiation of frequency 9.00 × 10^15 Hz is incident on a clean metal surface. The maximum kinetic energy of a photoelectron ejected from this surface is 5.70 × 10^-18... show full transcript

Worked Solution & Example Answer:Radiation of frequency 9.00 × 10^15 Hz is incident on a clean metal surface - Scottish Highers Physics - Question 12 - 2016

Step 1

Calculate the energy of the incident radiation

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Answer

The energy of the incident radiation can be calculated using the formula: E=hfE = h f where:

  • hh is Planck's constant, approximately 6.63×10−346.63 × 10^{-34} J·s,
  • ff is the frequency, which is given as 9.00×10159.00 × 10^{15} Hz.

Thus, the energy becomes: E=(6.63×10−34extJ⋅s)(9.00×1015extHz)=5.967×10−18extJE = (6.63 × 10^{-34} ext{ J·s})(9.00 × 10^{15} ext{ Hz}) = 5.967 × 10^{-18} ext{ J}

Step 2

Determine the work function of the metal

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Answer

The work function (heta heta) is found using the equation: K.E.=E−hetaK.E. = E - heta where:

  • K.E.K.E. is the maximum kinetic energy of the ejected photoelectron, which is given as 5.70×10−185.70 × 10^{-18} J.

Rearranging gives: heta=E−K.E.=5.967×10−18extJ−5.70×10−18extJ=0.267×10−18extJ=2.67×10−19extJ heta = E - K.E. = 5.967 × 10^{-18} ext{ J} - 5.70 × 10^{-18} ext{ J} = 0.267 × 10^{-18} ext{ J} = 2.67 × 10^{-19} ext{ J}

Therefore, the work function of the metal is 2.67×10−192.67 × 10^{-19} J, which corresponds to option A.

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