What transition in the hydrogen spectrum woul | vedclass

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(A) For $He^{+}$ ion,the wave number $(\bar{v})$ associated with the transition $n_2=4$ to $n_1=2$ is given by the Rydberg formula:$\bar{v} = \frac{1}

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$n=2$ to $n=1$

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(A) For $He^{+}$ ion,the wave number $(\bar{v})$ associated with the transition $n_2=4$ to $n_1=2$ is given by the Rydberg formula:$\bar{v} = \frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$For $He^{+}$,$Z=2$,$n_1=2$,and $n_2=4$:$\frac{1}{\lambda} = R(2)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 4R \left( \frac{1}{4} - \frac{1}{16} \right) = 4R \left( \frac{3}{16} \right) = \frac{3R}{4}$For the hydrogen atom,$Z=1$. Let the transition be from $n_2$ to $n_1$:$\frac{1}{\lambda} = R(1)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = \frac{3R}{4}$$\frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{3}{4}$Comparing this with the Lyman series formula where $n_1=1$:$\frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4}$Thus,the transition $n=2$ to $n=1$ in the hydrogen spectrum has the same wavelength.

List-$I$	List-$II$
$(1)$ Lyman	$(A)$ $n = 5 \to n = 4$
$(2)$ Paschen	$(B)$ $n = 3 \to n = 2$
$(3)$ Brackett	$(C)$ $n = 6 \to n = 5$
$(4)$ Balmer	$(D)$ $n = 4 \to n = 3$
	$(E)$ $n = 7 \to n = 6$

What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition $n=4$ to $n=2$ of $He^{+}$ spectrum?

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