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.

Options

A. `7.50 xx 10^(4)s^(-1)`
B. `4.76 xx10^(14)s^(-1)`
C. `3.15 xx 10^(15) s^(-1)`
D. `1.24 xx 10^(15) s^(-1)`

Answer

D

Explanation

Energy `= N_(A) hv`
`495.5 =6.023 xx 10^(23) xx 6.6 xx 10^(-34) xx v`
`v= (495.5 xx 10^(3)J)/(6.023 xx 10^(23) xx 6.6xx 10^(-34)) = 12.4 xx 10^(14)`
` = 1.24 xx 10^(15) s^(-1)`