STRUCTURE OF ATOM( NUMERICAL )

DAY 3
6.5.20
Good Morning Boys,
We will continue with the numerical practice of structure of atoms.

Learning Outcomes :
students will be able to

• Apply Formula
• understand concept
• Identify variables.

Note : Google form assessment practice on 8.5.20 .Chapter for Assessment will be structure of atom.

1. What is the maximum number of emission lines when the excited electron of an H atom in n = 6 drops to the ground state?

Answer:

When the excited electron of an H atom in n = 6 drops to the ground state, the following transitions are possible:

Hence, a total number of (5 + 4 + 3 + 2 + 1) 15 lines will be obtained in the emission spectrum.

The number of spectral lines produced when an electron in the n^th level drops down to the ground state is given by.

Given,

n = 6

Number of spectral lines = 15

Question 2.

(i) The energy associated with the first orbit in the hydrogen atom is –2.18 × 10^–18 J atom^–1. What is the energy associated with the fifth orbit?

(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.

Answer:

(i) Energy associated with the fifth orbit of hydrogen atom 

E₅ = –8.72 × 10^–20 J

(ii) Radius of Bohr’s n^th orbit for hydrogen atom is given by,

r_n = (0.0529 nm) n²

For,

n = 5

r₅ = (0.0529 nm) (5)²

r₅ = 1.3225 nm

Question 3.

Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

Answer:

For the Balmer series, n_i = 2. Thus, the expression of wavenumber is given by,

Wave number  is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition,  has to be the smallest.

For  to be minimum, n_f should be minimum. For the Balmer series, a transition from n_i = 2 to n_f = 3 is allowed. Hence, taking n_f = 3, we get:

 = 1.5236 × 10⁶ m^–1

Question 4

Calculate the wavelength of an electron moving with a velocity of 2.05 × 10⁷ ms^–1.

Answer:

According to de Broglie’s equation,

Where,

λ = wavelength of moving particle

m = mass of particle

v = velocity of particle

h = Planck’s constant

Substituting the values in the expression of λ:

Hence, the wavelength of the electron moving with a velocity of 2.05 × 10⁷ ms^–1 is 3.548 × 10^–11 m.

Question 5

The mass of an electron is 9.1 × 10^–31 kg. If its K.E. is 3.0 × 10^–25 J, calculate its wavelength.

Answer:

From de Broglie’s equation,

Given,

Kinetic energy (K.E) of the electron = 3.0 × 10^–25 J

Substituting the value in the expression of λ:

Hence, the wavelength of the electron is 8.9625 × 10^–7 m.

Question 6

Which of the following are isoelectronic species i.e., those having the same number of electrons?

Na⁺, K⁺, Mg²⁺, Ca²⁺, S^2–, Ar

Answer:

Isoelectronic species have the same number of electrons.

Hence, the following are isoelectronic species:

1) Na⁺ and Mg²⁺ (10 electrons each)

2) K⁺, Ca²⁺, S^2– and Ar (18 electrons each)

Question 7

(i) Write the electronic configurations of the following ions: (a) H^– (b) Na⁺ (c) O^2–(d) F^–

(ii) What are the atomic numbers of elements whose outermost electrons are represented by (a) 3s¹ (b) 2p³ and (c) 3p⁵?

(iii) Which atoms are indicated by the following configurations?

(a) [He] 2s¹ (b) [Ne] 3s² 3p³ (c) [Ar] 4s² 3d¹.

Answer:

(i) (a) H^– ion 

 Electronic configuration of H^– = 1s²

(b) Na⁺ 

 Electronic configuration of Na⁺ = 1s² 2s² 2p⁶ 3s⁰ or 1s² 2s² 2p⁶

(c) O^2– ion

 Electronic configuration of O^2– ion = 1s² 2s² p⁶

(d) F^– ion

∴ Electron configuration of F^– ion = 1s² 2s² 2p⁶

(ii) (a) 3s¹

Completing the electron configuration of the element as

1s² 2s² 2p⁶ 3s¹.

∴ Number of electrons present in the atom of the element

= 2 + 2 + 6 + 1 = 11

∴ Atomic number of the element = 11

(b) 2p³

Completing the electron configuration of the element as

1s² 2s² 2p³.

∴ Number of electrons present in the atom of the element = 2 + 2 + 3 = 7

∴ Atomic number of the element = 7

(c) 3p⁵

Completing the electron configuration of the element as

1s2 2s2 2p6 3s2 3p5∴ Number of electrons present in the atom of the element = 2 + 2 + 6+2+5 = 17

∴ Atomic number of the element = 17

(iii) (a) [He] 2s¹

The electronic configuration of the element is [He] 2s¹ = 1s² 2s¹.

∴ Atomic number of the element = 3

Hence, the element with the electronic configuration [He] 2s¹ is lithium (Li).

(b) [Ne] 3s² 3p³

The electronic configuration of the element is [Ne] 3s² 3p³= 1s² 2s² 2p⁶ 3s² 3p³.

∴ Atomic number of the element = 15

Hence, the element with the electronic configuration [Ne] 3s² 3p³ is phosphorus (P).

(c) [Ar] 4s² 3d¹

The electronic configuration of the element is [Ar] 4s² 3d¹= 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹.

∴ Atomic number of the element = 21

Hence, the element with the electronic configuration [Ar] 4s² 3d¹ is scandium (Sc).

Question 8

An electron is in one of the 3d orbitals. Give the possible values of n, l and m_l for this electron.

Answer:

For the 3d orbital:

Principal quantum number (n) = 3

Azimuthal quantum number (l) = 2

Magnetic quantum number (m_l) = – 2, – 1, 0, 1, 2

Question 9

What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?

Answer:

The n_i = 4 to n_f = 2 transition will give rise to a spectral line of the Balmer series. The energy involved in the transition is given by the relation,

Substituting the values in the given expression of E:

E = – (4.0875 × 10^–19 J)

The negative sign indicates the energy of emission.

Wavelength of light emitted 

Substituting the values in the given expression of λ:

Question 10

(i) An atomic orbital has n = 3. What are the possible values of l and m_l ?

(ii) List the quantum numbers (m_l and l) of electrons for 3d orbital.

(iii) Which of the following orbitals are possible?

1p, 2s, 2p and 3f

Answer:

(i) n = 3 (Given)

For a given value of n, l can have values from 0 to (n – 1).

∴ For n = 3

l = 0, 1, 2

For a given value of l, m_l can have (2l + 1) values.

For l = 0, m = 0

l = 1, m = – 1, 0, 1

l = 2, m = – 2, – 1, 0, 1, 2

∴ For n = 3

l = 0, 1, 2

m₀ = 0

m₁ = – 1, 0, 1

m₂ = – 2, – 1, 0, 1, 2

(ii) For 3d orbital, l = 2.

For a given value of l, m_lcan have (2l + 1) values i.e., 5 values.

∴ For l = 2

m₂ = – 2, – 1, 0, 1, 2

(iii) Among the given orbitals only 2s and 2p are possible. 1p and 3f cannot exist.

For p-orbital, l = 1.

For a given value of n, l can have values from zero to (n – 1).

∴ For l is equal to 1, the minimum value of n is 2.

Similarly,

For f-orbital, l = 4.

For l = 4, the minimum value of n is 5.

Hence, 1p and 3f do not exist.

THATS ALL FOR THE DAY.
MARK YOUR ATTENDANCE ON THE FORM.