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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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An ionic atom equivalent to hydrogen atom has wavelength equal to 1/4 of the wavelengths of hydrogen lines. The ion will be

${He}^{+}$

${Li}^{+ +}$

${Ne}^{9 +}$

${Na}^{10 +}$

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Explanation

(a) $\bar{ν} \propto \frac{1}{λ} \propto Z^{2} \Rightarrow {λZ}^{2} = constant \Rightarrow λ = \frac{λ}{4} Z^{2} \Rightarrow Z = 2$

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The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of 108.5 nm. The ground state energy of an electron of this ion will be

3.4 eV

13.6 eV

54.4 eV

122.4 eV

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Explanation

$∴ \frac{1}{λ} = {RZ}^{2} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$ gives $Z^{2} = \frac{n_{1}^{2} n_{2}^{2}}{(n_{2}^{2} - n_{1}^{2}) λR}$

On putting values Z = 2
From $E = - \frac{13.6 {(Z)}^{2}}{n^{2}} = \frac{- 13.6 (2)^{2}}{(1)^{2}} = - 54.4 eV$

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An electron in the n = 1 orbit of hydrogen atom is bound by 13.6 eV energy is required to ionize it is

13.6 eV

6.53 eV

5.4 eV

1.51 eV

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Explanation

(a) Ionization energy = Binding energy.

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Hydrogen atom emits blue light when it changes from n = 4 energy level to the n = 2 level. Which colour of light would the atom emit when it changes from the n = 5 level to the n = 2 level

Red

Yellow

Green

Violet

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Explanation

(d) In the transition from orbit $5 \to 2$ , more energy is liberated as compared to transition from $4 \to 2$

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The first line of Balmer series has wavelength 6563 Å. What will be the wavelength of the first member of Lyman series

1215.4 Å

2500 Å

7500 Å

600 Å

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Explanation

(a) $\frac{1}{λ_{Balmer}} = R [\frac{1}{1^{2}} - \frac{1}{2^{2}}] = \frac{3 R}{4} ∴ λ_{Lyman} = λ_{Balmer} \times \frac{5}{27} = 1215.4 \overset{0}{A}$

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In Bohr's model, the atomic radius of the first orbit is $r_{0}$ , then the radius of the third orbit is

$\frac{r_{0}}{9}$

$r_{0}$

$9 r_{0}$

$3 r_{0}$

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Explanation

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The wavelength of the energy emitted when electron come from fourth orbit to second orbit in hydrogen is 20.397 cm. The wavelength of energy for the same transition in ${He}^{+}$ is

5.099 ${cm}^{- 1}$

20.497 ${cm}^{- 1}$

40.994 ${cm}^{- 1}$

81.988 ${cm}^{- 1}$

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Explanation

(a) $E (= \frac{hc}{λ}) \propto \frac{Z^{2}}{n^{2}} \Rightarrow λ \propto \frac{1}{Z^{2}}$

Hence $λ_{{He}^{+}} = \frac{20.397}{4} = 5.099 cm$

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Minimum excitation potential of Bohr's first orbit in hydrogen atom is

13.6 V

3.4 V

10.2 V

3.6 V

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Explanation

(c) Excitation potential = $\frac{Excitation energy}{e}$
Minimum excitation energy corresponds to excitation from n =1 to n = 2
$∴$ Minimum excitation energy in hydrogen atom = -3.4-(-13.6) = +10.2 eV

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