QCE Vault / Physics Quantum theory — Question 89 Original QCE Vault practice · 5 marks
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QUESTION 89 (5 marks) Light of frequency 9.0 × 10 14 Hz 9.0\times10^{14}\,\text{Hz} 9.0 × 1 0 14 Hz illuminates a metal of work function 2.30 eV 2.30\,\text{eV} 2.30 eV . Determine the maximum kinetic energy of emitted photoelectrons in eV and the corresponding maximum electron speed. WORKED SOLUTION
Practice marking scheme 5 marks ANSWER E k , m a x = 1.42 eV E_{k,max}=1.42\,\text{eV} E k , ma x = 1.42 eV ; v m a x = 7.07 × 10 5 m s − 1 v_{max}=7.07\times10^5\,\text{m s}^{-1} v ma x = 7.07 × 1 0 5 m s − 1 . Worked solution
Photon energy:
E = h f = ( 6.626 × 10 − 34 ) ( 9.0 × 10 14 ) = 5.96 × 10 − 19 J = 3.72 eV . E=hf=(6.626\times10^{-34})(9.0\times10^{14})=5.96\times10^{-19}\,\text{J}=3.72\,\text{eV}. E = h f = ( 6.626 × 1 0 − 34 ) ( 9.0 × 1 0 14 ) = 5.96 × 1 0 − 19 J = 3.72 eV .
Hence
E k = 3.72 − 2.30 = 1.42 eV . E_k=3.72-2.30=1.42\,\text{eV}. E k = 3.72 − 2.30 = 1.42 eV .
In joules, E k = ( 1.42 ) ( 1.602 × 10 − 19 ) = 2.28 × 10 − 19 J E_k=(1.42)(1.602\times10^{-19})=2.28\times10^{-19}\,\text{J} E k = ( 1.42 ) ( 1.602 × 1 0 − 19 ) = 2.28 × 1 0 − 19 J .
Using E k = 1 2 m e v 2 E_k=\tfrac12m_ev^2 E k = 2 1 m e v 2 ,
v = 2 E k m e = 2 ( 2.28 × 10 − 19 ) 9.11 × 10 − 31 = 7.07 × 10 5 m s − 1 . v=\sqrt{\frac{2E_k}{m_e}}=\sqrt{\frac{2(2.28\times10^{-19})}{9.11\times10^{-31}}}=7.07\times10^5\,\text{m s}^{-1}. v = m e 2 E k = 9.11 × 1 0 − 31 2 ( 2.28 × 1 0 − 19 ) = 7.07 × 1 0 5 m s − 1 . Calculates photon energy.
[1 mark] Converts kinetic energy to joules.
[1 mark] Uses kinetic-energy relationship.
[1 mark] Calculates electron speed.
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