I&I Question (1 Viewer)

[MNsz]

The answer is very vague, and I don't get it. I thought the Stopping Voltage would still increase.

Can anyone explain?

 

[Fizzy_Cyst]

The stopping voltage would increase, but not necessarily double.
In the example in the question. Seeing as the Work function is 6x1014Hz, then the KEmax is 3x1014Hz and the stopping voltage occurs when qVstop = KEmax

i.e., Vstop = KEmax / q

If the frequency is doubled, to 1.8x1015Hz, then the KEmax would be 1.2x1015Hz.

Seeing as Vstop = KEmax / q

Then, seeing as KEmax increases 4x, therefore Vstop would increase 4x. Not double =)

 

[MNsz]

Thank you.
What you said make sense.

 

[Fizzy_Cyst]

No probs =)