Ionization energy of gaseous Na atoms is `495.5" kJ mol"^(-1)`. The lowest possible frequency of light that ionizes sodium atom is `(h=6.626 xx 10^(-34) J s, N_(A) = 6.022 xx 10^(23)" mol"^(-1))`

Options

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

Answer

D

Explanation

The ionization energy is given by `E=N_(A) hv`
`E=495.5xx10^(3)=6.022xx10^(23)xx6.626xx10^(-34)xx v, v=(495.10^(3))/(6.022xx10^(23)xx6.626xx10^(-34))=1.24xx10^(15)`