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Question

Find the value of wave number $$\left( \bar { V }  \right) $$ in term of Rydberg's constant ($$R_H$$), when transition of electron takes place between two levels of $${ He }^{ + }$$ ion, whose sum is 4 and difference is 2.


A
89RH
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B
329RH
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C
19RH
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D
none of these
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Solution

The correct option is B $$\frac{32}{9}R_H$$
Let take states be $$n_{1}$$ and $$n_{2}$$
$$\therefore n_{1}+n_{2}=4$$
$$n_{1}-n_{2}=2$$
$$\therefore  n_{1}=3\rightarrow n_{2}=1$$

$$\bar{\nu }=R_{H}\times z^{2} \left(\dfrac{1}{n_2^1}-\dfrac{1}{n_2^2}\right)$$

$$=R_{H}\times 4\left(1-\dfrac{1}{9}\right)$$     $$ (z$$ for $$He^{+}=2)$$

$$=R_{H}\times 4\times \dfrac{8}{9}$$

$$=\dfrac{32}{9}R_{H}$$

Chemistry

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