Question

A double charged lithium atom is equivalent to hydrogen whose atomic number is 3. The wavelength of required radiation for emitting electron from first to third Bohr orbit in $L{i}^{++}$ will be (Ionisation energy of hydrogen atom is 13.6 eV)

• A. $182.51\stackrel{\circ }{A}$
• B. $177.17\stackrel{\circ }{A}$
• C. $142.25\stackrel{\circ }{A}$
• D. $113.74\stackrel{\circ }{A}$

Answer 1

The correct option is D $113.74\stackrel{\circ }{A}$

Energy of a electron in nth orbit of a hydrogen like atom is given by
$E_n=-13.6\frac{Z^2}{n^2}eV$, and Z = 3 for Li
Required energy for said transition
$\Delta E=E_3-E_1=13.6Z^2(\frac{1}{1^2}-\frac{1}{3^2})=13.6\times 3^2\left[\frac{8}{9}\right]=108.8eV=108.8\times 1.6\times {10}^{-19}J$
Now using $\Delta E=\frac{hc}{\Delta E}\Rightarrow \lambda =\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{108.8\times 1.6\times {10}^{-19}}=0.11374\times {10}^{-7}m\Rightarrow \lambda =113.74\stackrel{\circ }{A}$