Suppose $3.1 \times 10^{- 18}$ J energy is needed by the interior of the human eye to see an object. How many photons of light of $λ$ = 400 nm will be needed to see the object?
$h = 6 .6 \times 10^{- 34}$ Js.

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answer is 6.

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Detailed Solution

The energy of one photon is calculated as:

$E = \frac{hC}{λ} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{400 \times 10^{- 9}} = 4.95 \times 10^{- 19} J$

The number of photons required:

$number of photons = \frac{3.1 \times 10^{- 18} J }{4.95 \times 10^{- 19}} = 6.3 ≃ 6$

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