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Atoms MCQs for NEET — Physics Questions with Answers

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In terms of Rydberg's constant R, the wave number of the first Balmer line is

$\frac{5 R}{36}$

$\frac{8 R}{9}$

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Explanation

(c) Wave number = $\frac{1}{λ} = R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
For first Balmer line $n_{1}$ = 2, $n_{2}$ =3
Wave number

$∴$ Wave number = $R (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = R (\frac{9 - 4}{9 \times 4}) = \frac{5 R}{36}$

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Which of the transitions in hydrogen atom emits a photon of lowest frequency (n = quantum number)

n = 2 to n = 1

n = 4 to n = 3

n = 3 to n = 1

n = 4 to n = 2

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Explanation

The frequency of the emitted photon is inversely proportional to the square of the principal quantum number (n). The transition from n=4 to n=3 involves the smallest energy difference and hence the lowest frequency photon emission.

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According to Bohr's theory, the expressions for the kinetic and potential energy of an electron revolving in an orbit is given respectively by

$+ \frac{e^{2}}{8 {πε}_{0} r}$ and $- \frac{e^{2}}{4 {πε}_{0} r}$

$+ \frac{8 {πε}_{0} e^{2}}{r}$ and $- \frac{4 {πε}_{0} e^{2}}{r}$

$- \frac{e^{2}}{8 {πε}_{0} r}$ and $- \frac{e^{2}}{4 {πε}_{0} r}$

$+ \frac{e^{2}}{8 {πε}_{0} r}$ and $+ \frac{e^{2}}{4 {πε}_{0} r}$

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Explanation

(a) $P . E . = - \frac{{ke}^{2}}{r} = - \frac{e^{2}}{4 {πε}_{0} r}; K . E = - \frac{1}{2} (PE) = \frac{e^{2}}{8 {πε}_{0} r}$

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Ratio of the wavelengths of first line of Lyman series and first line of Balmer series is

1: 3

27 : 5

5 : 27

4 : 9

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Explanation

(c) $\frac{1}{λ} = R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
For first line of Lymen series $n_{1}$ = 1 and $n_{2}$ = 2
For first line of Balmer series $n_{1}$ = 2 and $n_{2}$ = 3
So, $\frac{λ_{Lyman}}{λ_{Balmer}} = \frac{5}{27}$

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The velocity of an electron in the second orbit of sodium atom (atomic number = 11) is v. The velocity of an electron in its fifth orbit will be

$\frac{22}{5} v$

$\frac{5}{2} v$

$\frac{2}{5} v$

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Explanation

(d) $v_{n} \propto \frac{1}{n} \Rightarrow \frac{v_{5}}{v_{2}} = \frac{2}{5} \Rightarrow v_{5} = \frac{2}{5} v_{2} = \frac{2}{5} v$

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The absorption transitions between the first and the fourth energy states of hydrogen atom are 3. The emission transitions between these states will be

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Explanation

(d) By using $N_{E} = \frac{n (n - 1)}{2} \Rightarrow N_{E} = \frac{4 (4 - 1)}{2} = 6$

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The ratio of longest wavelength and the shortest wavelength observed in the five spectral series of emission spectrum of hydrogen is

$\frac{4}{3}$

$\frac{525}{376}$

$\frac{900}{11}$

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Explanation

(d) Shortest wavelength comes from $n_{1} = \infty$ to $n_{2} = 1$ and longest wavelength comes from $n_{1} = 6$ to $n_{2} = 5$ in the given case. Hence

$\frac{1}{λ_{\min}} = R (\frac{1}{1^{2}} - \frac{1}{\infty^{2}}) = R \frac{1}{λ_{\max}} = R (\frac{1}{5^{2}} - \frac{1}{6^{2}}) = R (\frac{36 - 25}{25 \times 36}) = \frac{11}{900} R$

$∴ \frac{λ_{\max}}{λ_{\min}} = \frac{900}{11}$

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In Bohr's model of hydrogen atom, let PE represents potential energy and TE the total energy. In going to a higher level

PE decreases, TE increases

PE increases, TE increases

PE decreases, TE decreases

PE increases, TE decreases

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Explanation

(b) As n increases P.E. increases and K.E. decreases.

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According to Bohr's model, the radius of the second orbit of helium atom is

0.53 Å

1.06 Å

2.12 Å

0.265 Å

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Explanation

(b) $r = \frac{n^{2}}{Z} (r_{0}); \Rightarrow r_{(n = 2)} = \frac{{(2)}^{2}}{2} \times 0.53 = 1.06$ Å

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The fact that photons carry energy was established by

Doppler's effect

Compton's effect

Bohr's theory

Diffraction of light

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Explanation

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