Question

If the wave number of the first line in the Balmer series of hydrogen atom is $15000c{m}^{-1}$ , the wave number of the first line of the Balmer series of $L{i}^{2+}$:

• A. $1.43\times {10}^{4}c{m}^{-1}$
• B. $1.66\times {10}^{9}c{m}^{-1}$
• C. $13.5\times {10}^{5}c{m}^{-1}$
• D. $1.35\times {10}^{5}c{m}^{-1}$

Answer 1

Answer: (D)

$1.35\times {10}^{5}c{m}^{-1}$

Balmer series wavelength for an atom-

$\frac{1}{\lambda }=Z^2{R}_H(\frac{1}{2^2}-\frac{1}{n^2})n=3,4,5$

For the first line in the Balmer series of hydrogen atom-
${\nu }_o=15000c{m}^{-1}$

For $L{i}^{2+}$ wave space number $=Z^2\times {\nu }_o$

$=Z^2\times 15000c{m}^{-1}$

$=(9\times 15000)c{m}^{-1}$

$=1.35\times {10}^{5}c{m}^{-1}$

Hence, option D is correct.