 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2012/paper-1/section-a/1/solution

K_{m a x} = h c / λ - Φ

to eliminate the work function:

h = \frac{( 2.3 - 0.9 ) e}{c [( 200 \times 1 0 ^{- 9} ) ^{- 1} - ( 258 \times 1 0 ^{- 9} ) ^{- 1} ]} = 6.66 \times 1 0^{- 34} J s .

Here

e

converts the energy difference from electronvolts to joules. The precision of the rounded stopping energies justifies about two significant figures,

6.7 \times 1 0^{- 34} J s

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