Question

# Wave number of spectral line for a given transition is $$x\ cm^{-1}$$ for $$He^{+}$$, then its value for $$Be^{3+}$$ (isoelectronic of $$He^{+}$$) for same transition is:

A
x4cm1
B
x cm1
C
4x cm1
D
16x cm1

Solution

## The correct option is D $$4x\ cm^{-1}$$$$\overline {v}$$ (wave number) $$= \overline{R_{H}} Z^{2}\left [\dfrac {1}{n_{1}^{2}} - \dfrac {1}{n_{2}^{2}}\right ]$$$$\overline {v_{1}} (He^{+}, Z = 2) = \overline {R_{H}}(2)^{2} \left [\dfrac {1}{n_{1}^{2}} - \dfrac {1}{n_{2}^{2}}\right ]$$$$\overline {v_{1}} (Be^{+}, Z = 4) = \overline {R_{H}}(4)^{2} \left [\dfrac {1}{n_{1}^{2}} - \dfrac {1}{n_{2}^{2}}\right ]$$$$\therefore \dfrac {v_{2}}{v_{1}} = \dfrac {(4)^{2}}{(2)^{2}}$$$$= \dfrac {16}{4} = 4$$$$\therefore \overline {v_{2}} = 4\overline {v_{1}}$$$$\overline {v_{2}} = 4\ x\ cm^{-1}$$Hence, the correct option is $$C$$Chemistry

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