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Electromagnetic rediations of wavelength `242 nm` is just sufficient to lonise Sodium atom. Then the ionisation energy of Sodium in kJ `mole^(-1)` is.

Text Solution

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`lamda=242nm=242xx10^(-9)m`
`C=3xx10^(8)ms^(-1)`
`E=hv=hxxC/(lamda)=6.6256xx10^(-34)xx(3xx10^(8))/(242xx10^(-9))`
`=0.082xx10^(-17)J=0.082xx10^(-20)xx6.02xx10^(23)`
`=493.6kJ "mol"^(-1)`
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