Question

Metal A has a higher work function than metal B.

The lowest frequency of light that will cause electrons to be emitted from metal B is fB

Which of the following is true:

Select one:
a. light of any frequency will cause photoelectrons to be emitted from metal A, but only above a threshold intensity determined by the work function
b. light of any frequency will cause photoelectrons to be emitted from metal A, but the lower the frequency, the longer it takes for the photoelectrons to start being emitted
c. light of frequency fB will cause photoelectrons to be emitted, but the maximum current will be less.
d. light of frequency fB will emit photoelectrons from metal A if the biasing voltage is increased enough
e. light of frequency higher than fB is needed emit photoelectrons from metal A, regardless of its intensity

Answer 1

Answer:

Metal A has a higher work function than metal B and the lowest frequency of light that will cause electrons to be emitted from metal B is f_B. If f_A be the frequency of light that will cause electrons to be emitted from metal A then the work function of metal A if W_A = hf_A.

Now, the work function of metal B will be W_B = hf_B.

According to the first sentence, W_A > W_B

Therefore, hf_A > hf_B or f_A > f_B.

Hence, correct answer is option (e) light of frequency higher than fB is needed emit photoelectrons from metal A, regardless of its intensity .