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## Homework Statement

**Suppose a filter allowed through only light of frequency fo (the cut-off frequency) or lower. In this case, what would the stopping potential be?**## Homework Equations

KEmax = (charge of electron)(stopping potential) = hf - Work function, where Work function = hfcutoff

## The Attempt at a Solution

Because the filter is now only allowing lights AT or BELOW the cutoff frequency, we can arrange our eaution as follows:

(e)(Vstopping) = hf - hfcutoff = h(f-fcutoff). But because f = fcutoff, (e)(Vstopping) = 0. Since e has a value of 1.6 x 10^-19 C, this forces stopping potential to be 0. However: this was marked incorrect.

Any suggestions or comments would be greatly appreciated!