Question

A doubly ionized Lithium atom is like a hydrogen atom with atomic number 3. What is the wavelength of the radiation required to excite the electron from first to third Bohr's orbit?

• A. $11114\stackrel{\circ }{A}$
• B. $114\stackrel{\circ }{A}$
• C. $14\stackrel{\circ }{A}$
• D. $4\stackrel{\circ }{A}$

Answer 1

The correct option is B $114\stackrel{\circ }{A}$

For Lithium atom, $Z=3$.

Let us first calculate the energy required to move from first to third Bohr's orbit.

From $E=-13.6\times \frac{Z^2}{n^2}$
$E_1$ = $-13.6\times \frac{3^2}{1}=-122.4eV$
$E_3$ = $-13.6\times \frac{3^2}{3^2}=-13.6eV$
Hence, the energy required to to move from first to third Bohr's orbit is
= $E_3-E_1$
= $122.4-13.6$
= $108.8eV$

Hence $108.8eV$ energy of photon is required for the excitement.
The wavelength associated with a photon can be calculated from the formula

$E=\frac{hc}{\lambda }$
i.e $108.8=\frac{12431}{\lambda \text{(}in\stackrel{\circ }{A})}$
Hence $\lambda =114A^o$