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Atoms NEET Questions — Free MCQ Practice

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The energy of electron in first excited state of H-atom is -3.4 eV its kinetic energy is

A.

– 3.4 eV

B.

+ 3.4 eV

C.

– 6.8 eV

D.

6.8 eV

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Explanation

(b) Kinetic energy = |Total energy|

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Which of the following phenomena suggests the presence of electron energy levels in atoms?

A.

Radio active decay

B.

Isotopes

C.

Spectral lines

D.

$α$ -particles scattering

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Explanation

(c) Spectral lines result from emission or absorption of energy due to difference in energy levels of electrons.

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Which of the following spectral series in hydrogen atom give spectral line of 4860 Å

A.

Lyman

B.

Balmer

C.

Paschen

D.

Brackett

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Explanation

(b) For Balmer series :

$λ_{m i n} = 3648 \overset{o}{A} λ_{m a x} = 6563 \overset{o}{A}$

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When an electron in hydrogen atom is excited, from its 4th to 5th stationary orbit, the change in angular momentum of electron is (Planck’s constant: h = $6.6 \times 10^{- 34} J - s$ 6.6)

A.

$4.16 \times 10^{- 34} J - s$

B.

$3.32 \times 10^{- 34} J - s$

C.

$1.05 \times 10^{- 34} J - s$

D.

$2.08 \times 10^{- 34} J - s$

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Explanation

(c) Change in the angular momentum

$∆ L = L_{2} - L_{1} = \frac{n_{2} h}{2 π} - \frac{n_{1} h}{2 π} \Rightarrow ∆ L = \frac{h}{2 π} (n_{2} - n_{1}) = \frac{6.6 \times 10^{- 34}}{2 \times 3.14} (5 - 4) = 1.05 \times 10^{- 34} J - s$

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The concept of stationary orbits was proposed by

A.

Neil Bohr

B.

J.J. Thomson

C.

Ruther ford

D.

I. Newton

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Explanation

(a) The concept of stationary orbits was proposed by Neil Bohr.

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The time of revolution of an electron around a nucleus of charge Ze in nth Bohr orbit is directly proportional to

A.

n

B.

$\frac{n^{3}}{Z^{2}}$

C.

$\frac{n^{2}}{Z}$

D.

$\frac{Z}{n}$

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Explanation

(b) $T = \frac{2 πr}{v}$ ; r = radius of nth orbit= $\frac{n^{2} h^{2}}{{πmZe}^{2}}$

v = speed of $e^{-}$ in nth orbit = $\frac{{ze}^{2}}{2 ε_{0} nh}$

$∴ T = \frac{4 ε_{0}^{2} n^{3} h^{3}}{{mZ}^{2} e^{4}} \Rightarrow T \propto \frac{n^{3}}{Z^{2}}$

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If R is the Rydberg’s constant for hydrogen the wave number of the first line in the Lyman series will be

A.

$\frac{R}{4}$

B.

$\frac{3 R}{4}$

C.

$\frac{R}{2}$

D.

2R

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Explanation

(b) For Lyman series

$\bar{v} = \frac{1}{λ} = R (\frac{1}{1^{2}} - \frac{1}{n^{2}})$ here n =2, 3, 4, 5......

For first line

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

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In hydrogen atom, if the difference in the energy of the electron in n =2 and n = 3 orbits is E, the ionization energy of hydrogen atom is

A.

13.2 E

B.

7.2 E

C.

5.6 E

D.

3.2 E

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Explanation

(b) Energy $E = K [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$ (K = constant)
$n_{1}$ = 2 and $n_{2}$ = 3, so $E = K [\frac{1}{2^{2}} - \frac{1}{3^{2}}] = K [\frac{5}{36}]$
For removing an electron $n_{1}$ = 1 to $n_{2}$ = $\infty$
Energy $E_{1} = K [1] = \frac{36}{5} E = 7.2 E$
$∴$ Ionization energy = 7.2 E

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The first member of the Paschen series in hydrogen spectrum is of wavelength 18,800 Å. The short wavelengths limit of Paschen series is

A.

1215 Å

B.

6560 Å

C.

8225 Å

D.

12850 Å

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Explanation

(c) For Paschen series $\bar{v} = \frac{1}{λ} = R [\frac{1}{3^{2}} - \frac{1}{n^{2}}]$ ; n=4, 5, 6....
For first member of Paschen series n = 4
$\frac{1}{λ_{1}} = R [\frac{1}{3^{2}} - \frac{1}{4^{2}}] \Rightarrow \frac{1}{λ_{1}} = \frac{7 R}{144}$

$\Rightarrow R = \frac{144}{7 λ_{1}} = \frac{144}{7 \times 18800 \times 10^{- 10}} = 1.1 \times 10^{- 7}$

For shortest wave length n = $\infty$
So

$\frac{1}{λ} = R [\frac{1}{3^{2}} - \frac{1}{\infty^{2}}] = \frac{R}{9} \Rightarrow λ = \frac{9}{R} = \frac{9}{1.1 \times 10^{- 7}} = 8.225 \times 10^{- 7} m = 8225 \overset{0}{A}$

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The ratio of the largest to shortest wavelengths in Lyman series of hydrogen spectra is

A.

$\frac{25}{9}$

B.

$\frac{17}{6}$

C.

$\frac{9}{5}$

D.

$\frac{4}{3}$

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Explanation

(d) For Lyman series $\frac{1}{λ_{\max}} = R [\frac{1}{1^{2}} - \frac{1}{2^{2}}] = \frac{3}{4} R$ and

$\frac{1}{λ_{\min}} = R [\frac{1}{1^{2}} - \frac{1}{\infty^{2}}] = \frac{R}{1} \Rightarrow \frac{λ_{\max}}{λ_{\min}} = \frac{4}{3}$

Atoms MCQs for NEET — Physics Questions with Answers

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