Page No 67: - Chapter 2 Structure Of Atom class 11 ncert solutions Chemistry - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 2.31:

How many electrons in an atom may have the following quantum numbers?(a) n = 4,   (b) n = 3, l = 0

Answer:

(a) Total number of electrons in an atom for a value of n = 2n²

∴ For n = 4,

Total number of electrons = 2 (4)²

= 32

The given element has a fully filled orbital as

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰.

Hence, all the electrons are paired.

∴ Number of electrons (having n = 4 and  ) = 16

(b) n = 3, l = 0 indicates that the electrons are present in the 3s orbital. Therefore, the number of electrons having n = 3 and l = 0 is 2.

Question 2.32:

Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.

Answer:

Since a hydrogen atom has only one electron, according to Bohr’s postulate, the angular momentum of that electron is given by:

Where,

n = 1, 2, 3, …

According to de Broglie’s equation:

Substituting the value of ‘mv’ from expression (2) in expression (1):

Since ‘2πr’ represents the circumference of the Bohr orbit (r), it is proved by equation (3) that the circumference of the Bohr orbit of the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron revolving around the orbit.

Question 2.33:

What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He⁺ spectrum?

Answer:

For He⁺ ion, the wave number  associated with the Balmer transition, n = 4 to n = 2 is given by:

Where,

n₁ = 2

n₂ = 4

Z = atomic number of helium

According to the Question, the desired transition for hydrogen will have the same wavelength as that of He⁺.

By hit and trail method, the equality given by equation (1) is true only when

n₁ = 1and n₂ = 2.

∴ The transition for n₂ = 2 to n = 1 in hydrogen spectrum would have the same wavelength as Balmer transition n = 4 to n = 2 of He⁺ spectrum.

Question 2.34:

Calculate the energy required for the process

The ionization energy for the H atom in the ground state is 2.18 ×10^–18 J atom^–1

Answer:

Energy associated with hydrogen-like species is given by,

For ground state of hydrogen atom,

For the given process,

An electron is removed from n = 1 to n = ∞.

∴ The energy required for the process 

Question 2.35:

If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.

Answer:

1 m = 100 cm

1 cm = 10^–2 m

Length of the scale = 20 cm

= 20 × 10^–2 m

Diameter of a carbon atom = 0.15 nm

= 0.15 × 10^–9 m

One carbon atom occupies 0.15 × 10^–9 m.

∴ Number of carbon atoms that can be placed in a straight line

Question 2.36:

2 × 10⁸ atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.

Answer:

Length of the given arrangement = 2.4 cm

Number of carbon atoms present = 2 × 10⁸

∴ Diameter of carbon atom

Question 2.37:

The diameter of zinc atom is .Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.

Answer:

(a) Radius of zinc atom 

(b) Length of the arrangement = 1.6 cm

= 1.6 × 10^–2 m

Diameter of zinc atom = 2.6 × 10^–10 m

∴ Number of zinc atoms present in the arrangement

Question 2.38:

A certain particle carries 2.5 × 10^–16C of static electric charge. Calculate the number of electrons present in it.

Answer:

Charge on one electron = 1.6022 × 10^–19 C

⇒ 1.6022 × 10^–19C charge is carried by 1 electron.

∴ Number of electrons carrying a charge of 2.5 × 10^–16 C

Question 2.39:

In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is –1.282 × 10^–18C, calculate the number of electrons present on it.

Answer:

Charge on the oil drop = 1.282 ×10^–18C

Charge on one electron = 1.6022 × 10^–19C

∴Number of electrons present on the oil drop

Question 2.40:

In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the α-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results?

Answer:

A thin foil of lighter atoms will not give the same results as given with the foil of heavier atoms.

Lighter atoms would be able to carry very little positive charge. Hence, they will not cause enough deflection of α-particles (positively charged).

Question 2.41:

Symbols  can be written, whereas symbols  are not acceptable. Answer briefly.

Answer:

The general convention of representing an element along with its atomic mass (A) and atomic number (Z) is 

Hence, is acceptable but is not acceptable.

can be written but  cannot be written because the atomic number of an element is constant, but the atomic mass of an element depends upon the relative abundance of its isotopes. Hence, it is necessary to mention the atomic mass of an element

Question 2.42:

An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.

Answer:

Let the number of protons in the element be x.

∴ Number of neutrons in the element

= x + 31.7% of x

= x + 0.317 x

= 1.317 x

According to the Question,

Mass number of the element = 81

∴ (Number of protons + number of neutrons) = 81

Hence, the number of protons in the element i.e., x is 35.

Since the atomic number of an atom is defined as the number of protons present in its nucleus, the atomic number of the given element is 35.

∴ The atomic symbol of the element is  .

Question 2.43:

An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion.

Answer:

Let the number of electrons in the ion carrying a negative charge be x.

Then,

Number of neutrons present

= x + 11.1% of x

= x + 0.111 x

= 1.111 x

Number of electrons in the neutral atom = (x – 1)

(When an ion carries a negative charge, it carries an extra electron)

∴ Number of protons in the neutral atom = x – 1

Given,

Mass number of the ion = 37

∴ (x – 1) + 1.111x = 37

2.111x = 38

x = 18

∴The symbol of the ion is 

Question 2.44:

An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion.

Answer:

Let the number of electrons present in ion 

∴ Number of neutrons in it = x + 30.4% of x = 1.304 x

Since the ion is tripositive,

⇒ Number of electrons in neutral atom = x + 3

∴ Number of protons in neutral atom = x + 3

Given,

Mass number of the ion = 56

∴ Number of protons = x + 3 = 23 + 3 = 26

∴ The symbol of the ion 

Question 2.45:

Arrange the following type of radiations in increasing order of frequency: (a) radiation from microwave oven (b) amber light from traffic signal (c) radiation from FM radio (d) cosmic rays from outer space and (e) X-rays.

Answer:

The increasing order of frequency is as follows:

Radiation from FM radio < amber light < radiation from microwave oven < X- rays < cosmic rays

The increasing order of wavelength is as follows:

Cosmic rays < X-rays < radiation from microwave ovens < amber light < radiation of FM radio

Question 2.46:

Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6 × 10²⁴, calculate the power of this laser.

Answer:

Power of laser = Energy with which it emits photons

Power 

Where,

N = number of photons emitted

h = Planck’s constant

c = velocity of radiation

λ = wavelength of radiation

Substituting the values in the given expression of Energy (E):

E 

= 0.3302 × 10⁷ J

= 3.33 × 10⁶ J

Hence, the power of the laser is 3.33 × 10⁶ J.

Question 2.47:

Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission, (b) distance traveled by this radiation in 30 s (c) energy of quantum and (d) number of quanta present if it produces 2 J of energy.

Answer:

Wavelength of radiation emitted = 616 nm = 616 × 10^–9 m (Given)

(a) Frequency of emission 

Where,

c = velocity of radiation

λ = wavelength of radiation

Substituting the values in the given expression of :

= 4.87 × 10⁸ × 10⁹ × 10^–3 s^–1

ν = 4.87 × 10¹⁴ s^–1

Frequency of emission (ν) = 4.87 × 10¹⁴ s^–1

(b) Velocity of radiation, (c) = 3.0 × 10⁸ ms^–1

Distance travelled by this radiation in 30 s

= (3.0 × 10⁸ ms^–1) (30 s)

= 9.0 × 10⁹ m

(c) Energy of quantum (E) = hν

(6.626 × 10^–34 Js) (4.87 × 10¹⁴ s^–1)

Energy of quantum (E) = 32.27 × 10^–20 J

(d) Energy of one photon (quantum) = 32.27 × 10^–20 J

Therefore, 32.27 × 10^–20 J of energy is present in 1 quantum.

Number of quanta in 2 J of energy

= 6.19 ×10¹⁸

= 6.2 ×10¹⁸

Question 2.48:

In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 × 10^–18 J from the radiations of 600 nm, calculate the number of photons received by the detector.

Answer:

From the expression of energy of one photon (E),

Where,

λ = wavelength of radiation

h = Planck’s constant

c = velocity of radiation

Substituting the values in the given expression of E:

E 

E = 3.313 × 10^–19 J

Energy of one photon = 3.313 × 10^–19 J

Number of photons received with 3.15 × 10^–18 J energy

= 9.5

≈ 10