Question

What electronic transition in $L{i}^{2+}$ produces the radiation of same wavelength as the first line in the Balmer's series of hydrogen spectrum?

• A. $n_2=3$ to $n_1=2$
• B. $n_2=6$ to $n_1=3$
• C. $n_2=9$ to $n_1=6$
• D. $n_2=9$ to $n_1=8$

Answer 1

Answer: (C)

$n_2=9$ to $n_1=6$

According to Bohr's Theory, the energy of an electron in $n^{th}$ orbit is:

$E_n=-R_H\left(\frac{Z^2}{n^2}\right)J$

where $R_H=2.18\times {10}^{-18}J$ and Z is the atomic number of an atom or ion having one electron.

In the atomic emission spectra of Hydrogen, the electron jumps from higher orbit/ energy state $n_2$ to lower energy state $n_1$ with the emission of radiation. The energy of the emitted radiation is equal to the difference between the energy of the two states involved:

$\Delta E=E_{n_1}-E_{n_2}$

Therefore, solving for first line of Balmer series, $n_1=2and{n}_2-n_1=1$ for Hydrogen, $Z=1$ energy of emitted radiation is:

$\Delta E=-R_H(\frac{1}{2^2}-\frac{1}{3^2})$

For ${Li}^{2+}$ ($Z=3$), energy of radiation is:

$\Delta E=-R_H{.3}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

For the two radiations to be of same wavelength, energy difference between transition states should be same. Hence,

$1^2(\frac{1}{2^2}-\frac{1}{3^2})=3^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Or,

$\frac{1}{2^2}=\frac{3^2}{n_1^2}and\frac{1}{3^2}=\frac{3^2}{n_2^2}$

Hence,

$n_1=6and{n}_2=9$ transition is the answer.