light: quantim model (1 Viewer)

[kractus]

Also, im also unsire why this is B:

 

[fx82au]

i cba reading the top question, but for the 2013 q, its b because increasing intensity increases the amount of photoelectrons emitted if and only if the threshold frequency has been met. If frequency of the incident light is not increased, there will be no increase in energy, by E=hf. Furthermore, , so if frequency is not increaded, then will remain the same.

im bad at explaining things, so i hope this helps

 

[wizzkids]

2013 Q.20
First we have to convert 450 nanometres light to a frequency, and mark this point on the horizontal axis of the graph.
450 nm corresponds to 6.7 x 10¹⁴ Hz, which is below the cut-off frequency for metal Y, but above the cut-off frequency for metal X.
We are also aware that a change in the intensity of light does not increase the maximum K.E. of photoelectrons, but it does increase the number of photoelectrons, provided the light photon carries more energy than the cut-off frequency.
OK, so now we can consider the answers.
(A) is wrong, because the number of photoelectrons from metal X would increase.
(B) is correct, because of the reasoning shown above.
(C) is wrong for many reasons, but suffice to say the number of photoelectrons from metal X would increase.
(D) is wrong, because 450nm light does not carry sufficient quantum of energy to liberate any photoelectrons from metal Y.