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Ionization energy of gaseous Na atoms is...

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))`

A

`7.50 xx 10^(4)s^(-1)`

B

`4.76 xx 10^(14)s^(-1)`

C

`3.15 xx 10^(15)s^(-1)`

D

`1.24 xx 10^(15)s^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
D
The ionization energy is given by `E=N_(A)hv`
`E=495.5 xx 10^(3) =6.022 xx 10^(23) xx 6.626xx10^(-34)xx v, v=(495.5xx10^(3))/(6.022 xx 10^(23)xx6.626xx10^(-34))=1.24 xx 10^(15)s^(-1)`
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