Question

If the ionisation potential of an atom is $122.4V$. the energy of recoil atom, in Joule, will be

• A. $0.44\times {10}^{-25}$
• B. $2.44\times {10}^{-25}$
• C. $1.44\times {10}^{-25}$
• D. $3.44\times {10}^{-25}$

Answer 1

The correct option is C $1.44\times {10}^{-25}$

$122.4V$ shows that it is a lithium atom.
$\frac{hc}{\lambda }=122.4eV$
Wavelength $=10.13nm$
De-Broglie's wavelength $=h/momentum$
Momentum $=\frac{6.62\times {10}^{-34}}{10.13\times {10}^{-9}}$$=65.35\times {10}^{-27}m/s$

$E=\sqrt{2\times mass\times momentum}$

Mass $=7\times (1.67\times {10}^{-27})kg$

Lithium has mass number 7.