Question

Ground state energy of $H$-atom is $(-E_1)$, the velocity of photoelectrons emitted when photon of energy $E_2$ strikes stationary ${Li}^{2+}$ ion in ground state will be:

• A. $v=\sqrt{\frac{2(E_2-E_1)}{m}}$
• B. $v=\sqrt{\frac{2(E_2+9E_1)}{m}}$
• C. $v=\sqrt{\frac{2(E_2-9E_1)}{m}}$
• D. $v=\sqrt{\frac{2(E_2-3E_1)}{m}}$

Answer 1

Answer: (C)

$v=\sqrt{\frac{2(E_2-9E_1)}{m}}$

We know c

So for lithium atom $E=E_1\times 3^2$ or $9E_1$

Energy conversation pinciple:

$\frac{1}{2}m{v}^2=E_2-9E_1$

$v^2=\frac{2(E_2-9E_1)}{m}$

$v=\sqrt{\frac{2(E_2-9E_1)}{m}}$