Question

In an isolated $L{i}^{2+}$, electron jumps from n $=2$ to n$=1$. What is the recoil momentum of the ion (in kg- m/s)?

• A. $5.44x10-27$
• B. $48.96x10-27$
• C. $16.32x10-27$
• D. $8.16x10-27$

Answer 1

The correct option is A $5.44x10-27$

energy of the photon emitted is given by the formula,

$E=E_1-E_2$

$E=13.6(\frac{1}{1}-\frac{1}{4})eV$

$=13.6\times 0.75=10.2eV$

momentum of the emitted photon,

$P=\frac{E}{C}$

$=\frac{10.2\times 1.6\times {10}^{-19}}{3\times {10}^8}$

$=5.44\times {10}^{-27}Ns$