Ultraviolet radiation of frequency $7.70 \times 10^{14} \: Hz$ is incident on the surface of a metal - Scottish Highers Physics - Question 10 - 2017
Question 10
Ultraviolet radiation of frequency $7.70 \times 10^{14} \: Hz$ is incident on the surface of a metal.
Photoelectrons are emitted from the surface of the metal.
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Worked Solution & Example Answer:Ultraviolet radiation of frequency $7.70 \times 10^{14} \: Hz$ is incident on the surface of a metal - Scottish Highers Physics - Question 10 - 2017
Step 1
Calculate the energy of the incoming photon
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Answer
The energy of the incoming photon can be calculated using the formula:
E=hf
where:
E is the energy of the photon
h is Planck's constant (6.63×10−34J⋅s)
f is the frequency (7.70×1014Hz)
Calculating:
E=(6.63×10−34J⋅s)(7.70×1014Hz)≈5.1×10−19J
Step 2
Relate maximum kinetic energy to work function
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Answer
The maximum kinetic energy (K.E.) of the emitted photoelectron is related to the work function (ϕ) by the equation:
K.E.=E−ϕ
Given:
K.E.=2.67×10−19J
E≈5.1×10−19J
We can rearrange the equation to solve for the work function:
ϕ=E−K.E.
Substituting the values:
ϕ=(5.1×10−19J)−(2.67×10−19J)≈2.44×10−19J
Step 3
Identify the work function from the options
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Answer
From our calculation, the work function is approximately 2.44×10−19J, which corresponds to option B.
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