Q. 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 $Li^{+ +}$ will be (Ionisation energy of hydrogen atom is $13.6eV$)
A) $182.51\ \mathring{\mathrm{A}}$
B) $177.17\ \mathring{\mathrm{A}}$
C) $142.25\ \mathring{\mathrm{A}}$
D) $113.74\ \mathring{\mathrm{A}}$
Explanation
(d) $E_{n} = - 13.6\frac{Z^{2}}{n^{2}}eV$. Required energy for said transition
$\mathrm{\Delta}E = E_{3} - E_{1} = 13.6\ Z^{2}\left\lbrack \frac{1}{1^{2}} - \frac{1}{3^{2}} \right\rbrack$
$\Rightarrow \mathrm{\Delta}E = 13.6 \times 3^{2}\left\lbrack \frac{8}{9} \right\rbrack = 108.8\ eV$
$\Rightarrow \mathrm{\Delta}E = 108.8 \times 1.6 \times 10^{- 19}J$
Now $\mathrm{\Delta}E = \frac{hc}{\lambda} = 108.8 \times 1.6 \times 10^{- 19}$
$\Rightarrow \lambda = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{108.8 \times 1.6 \times 10^{- 19}}$
$= 0.11374 \times 10^{- 7}m = 113.74\mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.