Electromagnetic radiation of wavelength `242` nm is just sufficient to ionise a sodium atom .Calculate the energy corresponding to tjhis wavelength and the ionisation potential of Na

Explanation

Gives `lambda = 242 nm = 242 xx 10^(-9)m`
From planck's theory , we have
`E = hv = (hc)/(lambda)`
or `E = (6.63 xx 3 xx 10^(8))/(242 xx 10^(-9)) J = (6.63 xx 3 xx 10^(-17))/(242)`
`= 8.21 xx 10^(-19) J`
Ionisation potential of Na
`= N_(A) xx E` (where `N_(A)` is Avogadro's number)
`= 6.022 xx 10^(23) xx 8.21 xx 10^(-19) J mol^(-1)`
`= 494.5 kJ mol^(-1)`