Light of frequency 2.0 × 10^{15} Hz is incident on a metal surface - AQA - A-Level Physics - Question 10 - 2019 - Paper 1
Question 10
Light of frequency 2.0 × 10^{15} Hz is incident on a metal surface. The work function of the metal is 4.6 × 10^{−19} J.
Which statement is correct?
A No photoelect... show full transcript
Worked Solution & Example Answer:Light of frequency 2.0 × 10^{15} Hz is incident on a metal surface - AQA - A-Level Physics - Question 10 - 2019 - Paper 1
Step 1
Calculate the energy of the incident photons
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Answer
The energy of the photons, E, can be calculated using the formula:
E=hf
where:
h is Planck's constant, approximately 6.63×10−34J⋅s
f is the frequency of the light, which is 2.0×1015Hz
So, substituting the values, we have:
E=(6.63×10−34J⋅s)(2.0×1015Hz)=1.326×10−18J
Step 2
Determine the maximum kinetic energy of the photoelectrons
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Answer
The maximum kinetic energy (KE) of the emitted photoelectrons is given by:
KE=E−W
where:
W is the work function of the metal, which is 4.6×10−19J
Substituting the values, we get:
KE=1.326×10−18J−4.6×10−19J=8.66×10−19J
This value rounds approximately to 8.7×10−19J. Hence, the correct answer is C.
Step 3
Identify the correct statement
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Answer
Based on the calculations above, the correct statement is:
C: Photoelectrons are released with a maximum kinetic energy of 8.7×10−19J.
