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Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom . Calculate the ionisation energy of sodium in kJ `mol^(-1)`.

A

`494.5xx10^(-6)` J/atom

B

`8169.5xx10^(-10)` J/atom

C

`5.85xx10^(-15)` J/atom

D

`8.214xx10^(-19)` J/atom

Text Solution

Verified by Experts

The correct Answer is:
D
`lambda=242nm=242xx10^(-9)m`
Energy required to ionise one atom of Na, `E=(hc)/lambda
= ((6.626xx10^(-34)Js)xx(3xx10^(8)ms^(-1)))/(242xx10^(-9)m)
=8.214xx10^(-19)J//atom`
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