Chemistry, asked by Cmmon2, 9 months ago

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

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Answered by Anonymous

$\huge\red{Answer}$

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Answered by DaDevilQueen

For He+ Ion, we have

$\huge \frac{1}{ \lambda} = {Z}^{2}R_{h}( \frac{1}{n_ {1}^{2} } - \frac{1}{n_ {2}^{2} } )$

$=( {2})^{2} R_h( \frac{1}{ {2}^{2} } - \frac{1}{ {4}^{2} } ) = R_h \frac{3}{4} .......(1)$

Now for hydrogen atom,

$\large \frac{1}{ \lambda} = R_h( \frac{1}{ {n_1}^{2} } - \frac{1}{n_2 ^{2} } ) \: ..........(2)$

Equating equation (1) and (2),we get.

$\huge \frac{1}{n_1 ^{2} } - \frac{1}{n_2 ^{2} } = \frac{3}{4}$

$\boxed { \red {\implies \: n_1 = 1 \: and \: n_2 = 2}}$

Hence, the transition n=2 to n=1 in hydrogen Atom will have the same wavelength as the transition n=4 to n=2 in He+ species.

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