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The energy of a photon of radiation having wavelength 300 nm is,A. `6.63xx10^(29)J`B. `6.63xx10^(-19)J`C. `6.63xx10^(-28)J`D. `6.63xx10^(-17)J` |
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Answer» Correct Answer - B `E=(hv)/(lamda)=(6.626xx10^(--34)xx3xx10^(8))/(300xx10^(-9))=6.63xx10^(-19)J` |
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