Ncert Exercises & Solutions

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

For ${He}^{+}$ ion, the wave number $(V)$ associated with the Balmer transition, n = 4 to n = 2 is given by:

$(v) = \frac{1}{λ} = {RZ}^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$

Where, $n_{1} = 2 n_{2} = 4$
Z = atomic number of helium

$(v) = \frac{1}{λ} = R {(2)}^{2} (\frac{1}{4} - \frac{1}{16})$
$= 4 R (\frac{4 - 1}{16})$
$v = \frac{1}{λ} = \frac{3 R}{4}$
$\Rightarrow λ = \frac{4}{3 R}$

According to the question, the desired transition for hydrogen will have the same wavelength as that of ${He}^{+}$ .

$\Rightarrow R (1^{2}) [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] = \frac{3 R}{4}$
$[\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] = \frac{3}{4} . . . (1)$

By hit and trail method, the equality given by equation (1) is true only when $n_{1} = 1 and n_{2} = 2$ .
$∴$ The transition for $n_{2}$ = 2 to n = 1 in hydrogen spectrum would have the same wavelength as Balmer transition n = 4 to n = 2 of ${He}^{+}$ spectrum.

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