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Photons of minimum energy `496k,J. mol^(-1)` are needed to an atoms. Calculate the lowest frequency of light that will ionize a sodium atom.A. `1.24xx10^(14)s^(-1)`B. `1.24xx10^(15)s^(-1)`C. `2.48xx10^(15)s^(-1)`D. `2.48xx10^(14)s^(-1)` |
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Answer» Correct Answer - 2 `Na(g)+496kJ mol^(-1)rarr Na^(+)(g)` For one photon `:E=(496kJ mol^(-1))/(N_(A)mol^(-1))=hv` `v=((496kJ mol^(-1))((10^(3)J)/(1kJ)))/((6.63xx10^(-34)Js)(6.022xx10^(23)mol^(-1)))=1.24xx10^(15)s^(-1)(Hz)` |
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