Question

hello
answer in notebook

dont copy

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Answer 1

$\sf \small \: ke = \frac{1}{2} \times 9 \times {10}^{ - 31} \times (6 \times {10}^{5} )^{2}$

$\sf \small161 = {10}^{ - 21} J \cong \: 1eV \:$

$\sf \small \: energy \: of \: photo \: E = \frac{12400}{4000} = 3.1eV \:$

$\sf \small \: E = ∅ + KE$

$\sf \small \:3.1 = ∅ + 1$

$\sf \small \: ∅ = 2.1 eV$

Answer 2

Explanation:

$\huge \tt ∅ \: = 2.1 \: ev$