Question

If the ionisation potential of an atom is $122.4V$. the recoil momentum of atom in Kg/m/sec will be

• A. $6.5\times {10}^{-26}$
• B. $1.8\times {10}^{-26}$
• C. $2.8\times {10}^{-26}$
• D. $3.8\times {10}^{-26}$

Answer 1

Answer: (A)

$6.5\times {10}^{-26}$

$E=122.4eV=\frac{1240}{\lambda }$
$\lambda =10.13nm$

$\therefore$ Recoil momentum will be equal to momentum given by the photon absorbed.

$p=\frac{h}{\lambda }$ $=\frac{6.6\times {10}^{-34}}{10.13\times {10}^{-9}}$$=0.65\times {10}^{-25}Kg/m/s$