The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 . The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is [2011]

Question

The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 $Å$ . The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is [2011]

Smart Achievers · Accepted Answer

(1)

We know for hydrogen or hydrogen-like atom,

$\frac{1}{λ} = R Z^{2} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$

For the first spectral line in the Balmer series of hydrogen atom, $n_{1} = 2$ and $n_{2} = 3$ . Here $Z = 1$

$∴$ $\frac{1}{6561} = R {(1)}^{2} (\frac{1}{4} - \frac{1}{9}) = \frac{5 R}{36}$ $\dots (i)$

For the second spectral line in the Balmer series of singly ionised helium ion, $n_{2} = 4$ and $n_{1} = 2$ $;$ $Z = 2$

$∴$ $\frac{1}{λ} = R {(2)}^{2} [\frac{1}{4} - \frac{1}{16}] = \frac{3 R}{4}$ $\dots (i i)$

Dividing Eq. (i) by Eq. (ii),

$\frac{λ}{6561} = \frac{5 R}{36} \times \frac{4}{3 R} = \frac{5}{27}$

$∴$ $λ = 1215 Å$

Solution

Q.
The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 $Å$ . The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is [2011]

1 1215 $Å$

2 1640 $Å$

3 2430 $Å$

4 4687 $Å$

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Solution

Q. The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Å. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is [2011]

1 1215 Å

2 1640 Å

3 2430 Å