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The energy of a photon is given as, `Delta E`/atom `= 3.03 xx 10^(-19)J "atom"^(-1)` then, the wavelength `(lambda)` of the photon isA. 6.56 nmB. 65.6 nmC. 656 nmD. 0.656 nm |
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Answer» Correct Answer - C (c) According to formula =, `E=(hc)/(lambda)(v=(c)/(lambda))` Energy E=hv `3.03xx10^(-19)=(hc)/(lambda)` `lambda=(6.63xx10^(-34)xx3.0xx10^(8))/(3.03xx10^(-19))` `=6.56xx10^(-7) m` `=6.56xx10^(-7)xx10^(9)nm` =`6.56xx10^(2) nm` `=656 nm` |
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