A photon of 300 nm is absorbed by a gas, which the re-emits two photons. One re-emitted photon has a wavelength of 400 nm. Calculate the energy of the other re-emitted photon.

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$1 .76 \times 10^{- 20} J$

$1.656 \times 10^{- 24} J$

$1 .656 \times 10^{- 19} J$

$1.892 \times 10^{- 20} J$

answer is A.

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Detailed Solution

$\begin{array}{ll} E_{total} & = \frac{hc}{λ} = \frac{6.626 \times 10^{- 34} Js \times 3 \times 10^{8} {ms}^{- 1}}{300 \times 10^{- 9} m} \\ = 6.626 \times 10^{- 19} J \\ E_{1} & = \frac{6.626 \times 10^{- 34} Js \times 3 \times 10^{8} {ms}^{- 1}}{400 \times 10^{- 9} m} \\ E_{2} & = (E_{total} - E_{1}) \\ = 6.626 \times 10^{- 19} \times (1 - \frac{3}{4}) = \frac{1}{4} \times 6.626 \times 10^{- 19} J \\ = 1.656 \times 10^{- 19} J \end{array}$