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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
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B
x cm1
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C
4x cm1
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D
16x cm1
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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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