Photoelectric Effect

[pnoybballin]

How does increasing the frequency of the incident photon impact the photoelectric cell in Figure 1?

A. More electrons will be ejected, but at the same speed
B. The same number of electrons will be ejected, but they will be faster
C. More electrons will be ejected, and they will be ejected with greater speed
D. The system will experience no change

The answer is B. But since E = hv / lamda and since v = ( f )( lambda) wouldn't the Energy decrease thus decreasing the speed of the electrons?

---

Members don't see this ad.

 

[BloodySurgeon]

E(in) = E(out). hf = hv/lamba. increase f and you will increase v. The number of electrons are not involved here therefore the answer is B.

 

[sleepy425]

The answer is B. But since E = hv / lamda and since v = ( f )( lambda) wouldn't the Energy decrease thus decreasing the speed of the electrons?

What? E=hv/lambda, v=(f)(lambda), so E=h(f)(lambda)/(lambda) so lambda cancels out and you get E=hf. So increasing f increases E. So each photon has higher energy, so the energy transferred to electrons in the metal is higher, so they have higher speed. Number of electrons doesn't change because frequency of the photon has nothing to do with the number of photons striking the metal, it just tells you about the energy of the photon.

 

[pnoybballin]

Niice. Makes perfect sense now. Thanks guys!