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|dw:1434346982679:dw|
oh no! it got uneven! :/ that is just one line for choice D!
here we have to apply this formula: \[\Large E = \frac{{hc}}{\lambda } = \frac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{730 \times {{10}^{ - 9}}}} = ...Joules\]
2.72465754E-19? choice A is the solution?
that's right!
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yay! thanks!:)
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