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The Lyman series of hydrogen spectrum lies in the region

Infrared

Visible

Ultraviolet

X- rays

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Explanation

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The size of an atom is of the order of

$10^{- 8}$ m

$10^{- 10} m$

$10^{- 12} m$

$10^{- 14} m$

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Explanation

(b) The size of the atom is of the order of 1Å = $10^{- 10}$ m.

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Which one of the series of hydrogen spectrum is in the visible region

Lyman series

Balmer series

Paschen series

Bracket series

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Explanation

(b) Balmer series lies in the visible region.

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The Rutherford $α$ -particle experiment shows that most of the $α$ -particles pass through almost unscattered while some are scattered through large angles. What information does it give about the structure of the atom

Atom is hollow

The whole mass of the atom is concentrated in a small centre called nucleus

Nucleus is positively charged

All the above

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Explanation

(d) According to Rutherford Atomic model, all of the options are correct.

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Which of the following is true

Lyman series is a continuous spectrum

Paschen series is a line spectrum in the infrared

Balmer series is a line spectrum in the ultraviolet

The spectral series formula can be derived from the Rutherford model of the hydrogen atom

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Explanation

(b) Paschen series lies in the infrared region.

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The energy required to knock out the electron in the third orbit of a hydrogen atom is equal to

13.6 eV

$+ \frac{13.6}{9} eV$

$- \frac{13.6}{3} eV$

$- \frac{3}{13.6} eV$

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Explanation

(b) Energy required to knock out the electron in the nth orbit $= + \frac{13.6}{n^{2}} eV \Rightarrow E_{3} = + \frac{13.6}{9} eV$

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An electron has a mass of $9.1 \times 10^{- 31} kg$ . It revolves round the nucleus in a circular orbit of radius $0.529 \times 10^{- 10}$ metre at a speed of $2.2 \times 10^{6} m / s$ . The magnitude of its linear momentum in this motion is

$1.1 \times 10^{- 34}$ kg-m/s

$2.0 \times 10^{- 24}$ kg-m/s

$4.0 \times 10^{- 24}$ kg-m/s

$4.0 \times 10^{- 31}$ kg-m/s

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Explanation

(b) Linear momentum = mv = $9.1 \times 10^{- 31} \times 2.2 \times 10^{6}$

= $2.0 \times 10^{- 24}$ kg-m/s

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The ionization potential for second He electron is

13.6 eV

27.2 eV

54.4 eV

100 eV

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Explanation

(c) For the ionization of second He electron. ${He}^{+}$ will act as hydrogen like atom.
Hence ionization potential $= Z^{2} \times 13.6 volt = {(2)}^{2} \times 13.6 = 54.4 V$

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The energy required to remove an electron in a hydrogen atom from n = 10 state is

13.6 eV

1.36 eV

0.136 eV

0.0136 eV

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Explanation

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Every series of hydrogen spectrum has an upper and lower limit in wavelength. The spectral series which has an upper limit of wavelength equal to 18752 Å is

Balmer series

Lyman series

Paschen series

Pfund series

(Rydberg constant R = $1.097 \times 10^{7}$ per metre)

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Explanation

(c) $\frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] \Rightarrow \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} = \frac{1}{Rλ} = \frac{1}{1.097 \times 10^{7} \times 18752 \times 10^{- 10}} = 0.0486 = \frac{7}{144}$

But $\frac{1}{3^{2}} - \frac{1}{4^{2}} = \frac{7}{144} \Rightarrow n_{1} = 3 and n_{2} = 4$ (Paschen series)

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