# Question Video: Understanding the Photoelectric Effect Physics • 9th Grade

In the photoelectric effect, which of the following should be done to increase the velocity of the emitted electrons? [A] Increasing the number of the incident photons that are of the critical frequency [B] Decreasing the number of the incident photons that are of the critical frequency [C] Increasing the frequency of the incident photons beyond the critical frequency [D] Decreasing the frequency of the incident photons below the critical frequency

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### Video Transcript

In the photoelectric effect, which of the following should be done to increase the velocity of the emitted electrons? (A) Increasing the number of the incident photons that are of the critical frequency. (B) Decreasing the number of the incident photons that are of the critical frequency. (C) Increasing the frequency of the incident photons beyond the critical frequency. (D) Decreasing the frequency of the incident photons below the critical frequency.

To begin, let’s recall a formula that can help us here. 𝐸 max equals ℎ𝑓 minus 𝑊, where 𝐸 max is the maximum kinetic energy of a photoelectron. ℎ is the Planck constant. 𝑓 is the frequency of the incident photon. And 𝑊 is the work function of a metal surface. This question is asking about how to increase the velocity of photoelectrons, and we know that their kinetic energy increases with their velocity. So our goal is to increase the value of this entire expression. Let’s think about how this could be done.

Well, we know that ℎ represents a constant, so there’s nothing we can do to increase its value. And for a given metal surface, the work function 𝑊 is also constant, so its value can’t really be modified either. This leaves us with one option: to increase the value of 𝑓, the frequency of the incident photons. This is a good hint that the correct answer is option (C). Option (D) also talks about the frequency of the incident photons. But since 𝐸 max is proportional to 𝑓, we want to increase the frequency, not decrease it. So let’s eliminate answer option (D).

Answer options (A) and (B) suggest another approach entirely: to change the number of incident photons. But all this could do is change the number of photoelectrons that are produced. As we saw before, the energy of the emitted electrons depends on the frequency of the incident photons. It doesn’t depend on the number of incident photons. Further, we should recall that the critical frequency, also known as the threshold frequency, gives the minimum frequency of an incident photon that can overcome the work function of a metal surface and induce the photoelectric effect. When an incident photon has this critical frequency, 𝐸 max is equal to zero since the energy of the incident photon is just enough to overcome the work function.

Therefore, if we want the emitted electrons to have greater energy and speed, we should definitely be using incident photons that have a greater frequency than the threshold frequency. Let’s eliminate options (A) and (B). Thus, we know the correct answer is option (C). Increasing the frequency of the incident photons beyond the critical frequency will increase the velocity of emitted electrons.