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A learner is investigating the photoelectric effect for two different metals, silver and sodium, using light of different frequencies - NSC Physical Sciences - Question 10 - 2016 - Paper 1
Question 10
A learner is investigating the photoelectric effect for two different metals, silver and sodium, using light of different frequencies. The maximum kinetic energy of ... show full transcript
Worked Solution & Example Answer:A learner is investigating the photoelectric effect for two different metals, silver and sodium, using light of different frequencies - NSC Physical Sciences - Question 10 - 2016 - Paper 1
Step 1
Define the term threshold frequency.
Answer
The threshold frequency is the minimum frequency of a photon (or light) required to emit electrons from the surface of a metal. At this frequency, the energy of the incoming photons is equal to the work function of the metal, allowing for the release of electrons.
Step 2
Which metal, sodium or silver, has the larger work function? Explain the answer.
Answer
Silver has the larger work function compared to sodium. This is evidenced by the slope of the graph representing each metal's photoelectric effect. A greater threshold frequency for silver indicates a higher work function, because a higher frequency photon is required to eject electrons from silver compared to sodium.
Step 3
Step 4
If light of the same frequency is shone on each of the metals, in which metal will the ejected photoelectrons have a larger maximum kinetic energy?
Answer
If light of the same frequency is shone on both metals, the ejected photoelectrons will have a larger maximum kinetic energy in sodium. This is because sodium's threshold frequency is lower, meaning it requires less energy to release electrons, thus providing more kinetic energy to the emitted electrons at the same frequency.
Step 5
Calculate the number of photons that will be incident on the metal plate per second.
Answer
First, we calculate the energy emitted by the bulb:
Next, we convert the wavelength of the blue light to frequency:
ightarrow f \\ ightarrow 6.38 imes 10^{14} ext{ Hz}$$ Now we calculate the energy of a single photon: $$E_{photon} = h f = (6.626 imes 10^{-34} ext{ J s})(6.38 imes 10^{14} ext{ Hz}) = 4.23 imes 10^{-19} ext{ J}$$ Finally, we find the total number of photons incident per second: $$ ext{Total photons} = rac{E_{bulb}}{E_{photon}} = rac{3 imes 10^{-3} ext{ W}}{4.23 imes 10^{-19} ext{ J}} = 7.09 imes 10^{15} ext{ photons/second}$$Step 6
Without any further calculation, write down the number of electrons emitted per second from the metal plate.
Answer
The number of electrons emitted per second from the metal plate will be the same as the number of photons incident per second, which is approximately 7.09 x 10^{15} electrons per second.