Structure of Atom Question Answers: NCERT Class 11 Chemistry

(i) Mass of an electron = 9.1 x 10^-28 g

Or

9.1 x 10^-28 g contains = 1 electron

Therefore 1g contains = 1/9.1 x 10^-28 *1 = 1.098 x 10²⁷ electrons

(ii) We know, one mole of electron = 6.022 x 10²³ electron

Mass of one electron = 9.1 × 10^–28 g

Or

mass of 6.022 x 10²³ electron = 9.1 x 10^-28 x 6.022 x 10²³ = 5.48 x 10^-4 g

Charge on one electron = 1.6 × 10^–19 coulomb

Charge on one mole of electron = (1.6 × 10^–19 C x 6.022 × 10²³)

= 9.63 × 10⁴ C

From E= hv or hc/ λ, we have λ = 242nm or 242 x 10^-9m

C = 3 x 10⁸ m/s

h = 6.62 x 10^-34 Js

Now putting these values in the equation we get

E= (6.62 x 10^-34 Js x 3 x 10⁸ m/s) / (242 x 10^-9m)

= 0.0821 x 10^-17 J/atom

Or

E = (0.0821 x 10^-17 ) / (1000 x 6.02 x 10²³)

= 494 KJ/mol

(Given)Power of bulb, P = 25 Watt = 25 Js^–1

We know, Energy of one photon, E = hν λ

Substituting the values in the given expression of E:

E = 34.87 × 10^–20 J

Rate of emission of quanta per second is given by R = P/ E, Where R is the rate of emission, P is the power & E is the energy

Substituting the values in the equation we get

Given λ = 6800 amgstrom or 6800x 10^-10 m = 6.8 x 10^-7m

C = 3 x 10⁸ m/s

Also λ = c/v or v = c/ λ Therefore putting the values in the equation, we get

=3 x 10⁸ / 6.8 x 10^-7m = 4.41 x 1014 s^-1

Hence, work function (W₀) of the metal = hν₀

= (6.626 × 10^–34 Js) (4.41 × 10¹⁴ s^–1)

= 2.922 × 10^–19 J

According to Balmer formula

Wave no = R_H [1/n₁² – 1/n₂²]

Here n1 = 2 ,       n2 = 4,     RH = 109678

Putting these values in the equation we get

Wave number =109678(1/2² - ¼²) = 109678 x 3/16

Also λ = 1/ wave number

Therefore λ = 16 / 109678 x 3 = 486 nm

The expression of energy is given by,

Where,

Z = atomic number of the atom

n = principal quantum number

For ionization from n1 = 5 to ,

Therefore ΔE = E2- E1 = - 21.8 X10^-19 (1/n₂²-1/n₁²)

= 21.8 X10^-19(1/n₂²-1/n₁²)

= 21.8 X10^-19 (1/5²-1/∞)

= 8.72 x 10^-20 J

For ionization from 1^st orbit, n₁= 1,

Therefore ΔE’ = 21.8x10^-19(1/1²-1/∞)

= 21.8x10^-19 J

Now ΔE’/ ΔE = 21.8x10^-19 / 8.72x10^-20 = 25

Thus the energy required to remove electron from 1^st orbit is 25 times than the required to electron from 5^th orbit.

Number of lines produced when electron from nth shell drops to ground state = n(n-1)/2

Now according to question lines produced when electron drops from 6^th shell to ground state

= 6(6-1) / 2 = 15

These are produced due to following transition

6 to 5      5 to 4    4 to 3     3to 2     2 to 1 (5 lines)

6 to 4     5 to 3    4 to 2     3 to 1             (4 lines)  

6 to 3     5 to 2    4 to 1                          (3 lines)

6 to 2     5 to 1                                      (2 lines)

6 to 1                                                    (1 lines)

Energy of an electrons = -(2.18x 10^-18)/n²

Where n= principal quantum number

Now the Energy associated with the fifth orbit of hydrogen atom is

\begin{align}E_{5}=\frac{-(2.18\times10^{-18})}{(5)^{2}}=\frac{-2.18\times10^{-18}}{25}\end{align}

                                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

According to Balmer formula

ṽ=1/λ = R_H[1/n₁²-1/n₂²]

For the Balmer series, n_i = 2.

Thus, the expression of wavenumber(ṽ) is given by,

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

For ṽ 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:

ṽ= 1.5236 × 10⁶ m^–1

Ground state energy (E₁) = – 2.18 × 10^–11 ergs (given)

 = –2.18 × 10^–11 × 10^–7 J

 = – 2.18 × 10^–18 J

Also the energy required to shift the electron from n = 1 to n = 5 is given as ΔE = E₅ – E₁

The expression for the energy of hydrogen of electron is

En = -2π²me⁴Z²/n²h²

Where m= mass of electrons

Z=atomic mass of atom 

e = charge of electron

h = planck’s constant

Now putting the values in the equation ΔE = E₅ – E₁ we get

The expression for the energy of hydrogen of electron is

E_n = -2π²me⁴Z²/n²h²

Where m= mass of electrons

Z=atomic mass of atom

e = charge of electron

h = planck’s constant

When n = 1 then E_n = - (2.18X10^-18 ) (GIVEN)

When n = 2 then En= - (2.18X10^-18 )/4 = 0.5465X10^-18 J

Therefore the energy required for ionization from n = 2 is 5.45 x 10^-19 J

Now wavelength of light needed

= E=hv = hc/λ

Or

λ = [{(6.62x10^-24)(3x10⁸)} / 5.45] x 10^-19 = 3647

(i) Molecule of CH⁴ (methane) contains electron = 10
Therefore 1 mole (6.022 x 10²³ atoms) contains electron = 6.022 x 10²⁴

(ii)  a) 1g atom of 14C = 14g = 6.022 x 10²³ atoms = 6.022 x 10²⁴ x 8 neutrons
Thus 14g or 14000 mg have 6.022 x 10²⁴ x 8 neutrons
Therefore 7 mg will have neutrons = 6.022 x 10²⁴ x 8 / 14000 x 7 = 2.4088 x 10²²

b) mass of 1 neutron = 1.675 x 10^-27 kg
Therefore mass of 2.4088 x 10²¹ neutrons = 2.4088 x 10²¹ x 1.67 x 10^-27 = 4.0347 x 10^-6 kg

(iii)  a) 1 mol of NH₃ = 17g NH₃ = 6.022 x 10²³ molecules of NH₃ = (6.022x10²³)(7 + 3) proton = 6.022 x 10²⁴ protons
Therefore 34 mg i.e 0.034 g NH₃ = 6.022 x 10²⁴ x 0.034/1 = 1.2044 x 10²² protons

b) mass of 1 proton = 1.6726 x 10^-27 kg
Therefore mass of 1.2044 x 10²² protons = (1.6726 x 10^-27)(1.2044 x 10²²) kg = 2.0145 x 10^-5 kg

No, the answer will not change with change in temperature & pressure.

V= 2.05 x10⁷ (given)

According to de Broglie’s equation,

Where, λ = wavelength of moving particle

m = mass of particle (9.10939 x 10^-31)

v = velocity of particle

h = Planck’s constant (6.62 x 10^-34)

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.

We know , KE = ½ mv²

Or

v = (2KE/M)^1/2

Or

v = 811.579 m/s

Now putting the values in the de Broglie’s equation

We get

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

Isoelectronic species = they are the species belonging to different atoms or ions which have same number of electrons but different magnitude of nuclear charge.

Now, A positive charge denotes the loss of an electron & A negative charge denotes the gain of an electron by a species.

1) Number of electrons in sodium (Na) = 11

Therefore, Number of electrons in (Na⁺) = 10

2) Number of electrons in K⁺ = 18

3) Number of electrons in Mg²⁺ = 10

4) Number of electrons in Ca²⁺ = 18

5) Number of electrons in sulphur (S) = 16

∴ Number of electrons in S^2- = 18

6) Number of electrons in argon (Ar) = 18

Hence, the following are isoelectronic species:

1) Na⁺ and Mg²⁺ (10 electrons each)

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

Electronic configuration of an atom is defined as the representation of the position of electrons in the various energy shell & subshells.

Now A negative charge on the species indicates the gain of an electron by it & A positive charge denotes the loss of an electron

(i) (a) H^– ion

The electronic configuration of H atom is 1s¹.(atomic number = 1)

∴ Electronic configuration of H^– = 1s²

(b) Na⁺ ion

The electronic configuration of Na atom is 1s² 2s² 2p⁶ 3s¹.(atomic number = 11)

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

(c) O^2– ion

The electronic configuration of 0 atom is 1s² 2s² 2p⁴.(atomic number = 8)

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

(d) F^– ion

The electronic configuration of F atom is 1s² 2s² 2p⁵.(atomic number = 9)

∴ 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(sodium)

(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(nitrogen)

(c) 3p⁵

Completing the electron configuration of the element as 1s² 2s² 2p⁵.

∴ Number of electrons present in the atom of the element = 2 + 2 + 5 = 9

∴ Atomic number of the element = 9(fluorine)

(iii) (a) [He] 2s1

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

∴ Atomic number of the element = 3 (lithium , a p-block element)

(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(phosphorous, a p block element)

(c) [Ar] 4s² 3d¹

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

∴ Atomic number of the element = 21(scandium , a d block element)

Quantum numbers = the set of four number which gives a complete information about the electrons in an atom.

Here n = principal quantum number

As for any value ‘n’ of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n – 1).

For n = 1 (K shell) has l = 0 (one subshell)

For n = 2 (L shell) has l= 0,1 (2 subshell)

For n = 3 (M shell) has l = 0,1,2 (3 subshell)

For n = 4 (N shell) has l = 0,1,2,3 (4 subshell)

For n = 5 (O shell) has l = 0,1,2,3,4 (5subshell)

∴ For l = 4, minimum value of n = 5

When l = 2 the value of m = -2,-1, 0 ,+1,+2

Now, for the 3d orbital:

Principal quantum number (n) = 3

Azimuthal quantum number (l) = 2

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

An atom consists of protons, electrons & neutrons.A positive atom will contain higher number of proton(H⁺), a negative atom will have higher number of electrons(e^-) & a neutral atom will contain same number of protons & electrons.Therefore,

(i) For an atom to be neutral, the number of protons is equal to the number of electrons.

∴ Number of protons in the atom of the given element = 29

(ii) The electronic configuration of the atom is

1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰. Which is the electronic configuration of copper 

Now A negative charge on the species indicates the gain of an electron by it & A positive charge denotes the loss of an electron

: Number of electrons present in hydrogen molecule (H2) = 1 + 1 = 2

∴ Number of electrons in  = 2 – 1 = 1

H2: Number of electrons in H2 = 1 + 1 = 2

: Number of electrons present in oxygen molecule (O₂) = 8 + 8 = 16

∴ Number of electrons in = 16 – 1 = 15

Here n= principal quantum number, l= azimuthal quantum number

(a) when n = 1, l = 0 (Given) The orbital is 1s. (can have maximum of 2 electron)

(b) For n = 3 and l = 1 The orbital is 3p. (can have maximum of 6 electrons).

(c) For n = 4 and l = 2 The orbital is 4d. (can have maximum of 10 electrons)

(d) For n = 4 and l = 3 The orbital is 4f. (can have maximum of 14 electrons)

We know atomic number(z) = no of protons = no of electrons

Also mass number(A) = no of protons + no of neutrons

¹³₆C:

Here mass number = 13

Atomic number = Number of protons = 6

Number of neutrons = (mass number) – (Atomic number)

= 13 – 6 = 7

:

Here mass number = 16

Atomic number = 8 Number of protons = 8 Number of neutrons = (mass number) – (Atomic number)

= 16 – 8 = 8

:

Here Mass number = 24

Atomic number = Number of protons = 12

Number of neutrons = (mass number) – (Atomic number)

= 24 – 12 = 12

:

Here mass number = 56

Atomic number = Number of protons = 26

Number of neutrons = (mass number ) – (Atomic number)

= 56 – 26 = 30

:

Here mass number = 88

Atomic number = Number of protons = 38

Number of neutrons = (mass number) – (Atomic number)

= 88 – 38 = 50

(a) The given set of quantum numbers is not possible because the value of the principal quantum number (n) cannot be zero.

(b) The given set of quantum numbers is possible.

(c) The given set of quantum numbers is not possible when n = 1, l is not equal to 1

(d) The given set of quantum numbers is possible.

(e) The given set of quantum numbers is not possible when n = 3 ,l = is not equal to 3

(f) The given set of quantum numbers is possible.

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

a) Total no of electrons when n=4 ,then n= 2x4² = 32 & half of them have m_s = -1/2

b) When n= 3, l=0 means 3s orbital which can have 2 electrons .

The angular momentum of an electron = mvr = nh/2π --------------1

Also according to de Broglie equation λ = h/mv

Or

mv = h/v -----------2

Putting 2 in 1

h/v =nh/2π or 2λr= nλ

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.

For an atom = 1/λ = R_HZ² (1/n₁²-1/n₂²)

For He⁺ spectrum Z = 2,  n₂=4,  n₁= 2

Therefore = 1/λ = R_H x 4(1/2²-1/4²) = 3R_H/4

For hydrogen spectrum = 3R_H/4,  Z = 1

Therefore = 1/λ = R_H X 1(1/n₁²-1/n₂²)

Or

R_H (1/n₁²-1/n₂²) = 3R_H/4

Or

1/n₁²-1/n₂²  = 3/4

Which can be so for n₁=1 & n₂ = 2, i.e the transition is from n = 2 to n=1

For H like particles,

E_n = - 2π²mZ²e⁴ / n²h²

For H atom  I.E = E – E₁ = 0 – (- 2π²me⁴ / 1²h²) = 2π²me⁴ / h²

= 2.18x 10^-18J /atom (given)

For the given process, energy required = E_n – E₁

= 0 -  ( - 2π²me⁴ / 1²h²)

= 4 x 2π²me⁴ / h²

= 4 x 2.18 x 10^-18J

= 8.72 x 10^-18 J

Diameter of carbon atom = 0.15nm = 0.15 x 10^-9m = 1.5 x 10^-10m

Length along which atoms are to be placed = 20cm = 20x10^-2m = 2 x 10^-1m

Therefore no of carbon atoms which can be placed along the length

= (2 x 10^-1m) / (1.5x10^-10 )

=1.33x10⁹

Total Length = 2.4 cm

Total number of atoms along the length = 2 × 10⁸

Therefore diameter of each atom = 2.4 / (2 × 10⁸)

= 1.2 x 10^-8 cm

And radius of the atom = (1.2 x 10^-8) / 2

= 0.60 x 10^-8cm

= 0.060 x 10^-9m

= 0.060nm

(a) radius = 2.6/2 = 1.3Å = 1.3 x 10^-10m = 130pm

(b) given 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

Charge carried by one electron = 1.6022 × 10^–19 C

Therefore electrons present in particle carrying 2.5x10^-16C charge =

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

Charge carried by one electron = 1.6022 × 10^–19C

∴electrons present on the oil drop carrying 1.282x10^-18C charge

1) the element with atomic number(Z) 17 & mass number (A) 35 is chlorine =

2) the element with atomic number(Z) 92 & mass number (A)233 is uranium =

3) the element with atomic number(Z) 4 & mass number (A) 9 is berellium =

In 1911, Rutherford performed alpha rays scattering experiment to demonstrate the structure of atom. Heavy atoms have a heavy nucleus carrying a large amount of positive charge.Hence,some alpha particles are easily deflected back on hitting the nucleus.Also a number of alpha particles are deflected through small angles because of large positive charge on the nucleus.If light atoms are use,their nuclei will be light & moreover,they will have small positive charge on the nucleus.Hence, the number of particles deflected back & those deflecte through some angle will be negligible.

The general convention of representing an element along with its atomic mass (A) and atomic number (Z) is Atomic number of an element is fixed.However,mass number is not fixed as it depends upon the isotope taken.Hence it is essential to indicate mass number

Mass number = 81 i.e p + n = 81

Let us suppose number of protons = x and neutrons

= x+31.7Xx /100 = 1.317x

Therefore x + 1.317x = 81

Or

2.317x = 81

Or x = 81/2.317 = 35 T

hus number of protons = 35 ,i.e atomic number = 35

Hence the symbol is ⁸¹₃₅ Br

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

Therefore 37 = 1.111x + x -1

Or 2.111 x = 38

Or x = 18

Therefore no of protons = atomic no = x -1 = 18 – 1 = 17

∴The symbol of the ion is 

Let the number of electrons present in ion ,S³⁺ = x

∴ 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

Now mass number = no of protons + no of neutrons

56 = x + 3 + 1.304x

Or 2.304x = 53

Or x = 23

Therefore no of protons = atomic no = x + 3 = 23 + 3 = 26

∴ The symbol of the ion 

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

E = Nhv = Nhc/λ

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):

= 3.33 × 10⁶ J

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

λ = 616nm  or = 616x10^-9m

a) Frequency , v=c/ λ = 3.8x10⁸ / 616x10^-9 = 4.87x10¹⁴ s^-1

b) Velocity of the radiation = 3x10⁸ m/s

Therefore distance travelled in 30s = 30x3x10⁸ = 9.0x10⁹m

c) E = hv = hc/ λ = (6.626x10^-34)(3.0x10⁸) / (616x10^-9 ) = 32.27x10^-20J

d) No of quanta in 2J of energy = 2/32.27 x 10^-28 = 6.2x10¹⁸

Energy of 1 photon = hv= hc/ λ = (6.626x10^-34)(3x10⁸) / 600x10^-9 = 3.313 x 10^-19J

Total energy received = 3.15 x10^-8J

Therefore no of photons received = 3.15x10^-18 / 3.313x10^-19 = 9.51

Frequency = 1/2.x10^-9 = 0.5x10⁹ / sec

Energy = Nhv = (2.5x10⁵)(6.626x10^-34)(0.5x10⁹)

= 8.28x10^-10J

Here λ = 580 nm(given)

We have to find out ν & 

Therefore from the equation  we can write,

Where, ν = frequency of yellow light

c = velocity of light in vacuum = 3 × 10⁸ m/s

λ = wavelength of yellow light = 580 nm = 580 × 10^–9 m

Substituting the values in we get

Thus, frequency of yellow light emitted from the sodium lamp

= 5.17 × 1014 s–1

Wave number = it is defined as the number of wavelengths which can be accommodated in 1cm length along the direction of propagation.

Wave number of yellow light is given by the equation

We have λ = 580 nm(given),putting the value in the equation we get

λ₁ = 589 nm = 5.89X10^-9m

therefore v₁=c/ λ₁ = 3x10⁸/589x10^-9 = 5.093x10¹⁴/sec

λ₂ = 589.6nm = 589.6x10^-9m

therefore v₂ = c/ λ₂ = 3x10⁸/589.6x10^-9

= 5.088x10¹⁴/sec

ΔE=E₂-E₁ = h(v₂-v₁)

= (6.626x10^-34)(5.093x-5.088)x10¹⁴

= 3.31x10^-22J

A) Work function of caesium (WO) = hv_o

Therefore v_o = W_O/h = 1.9 x1.602 x10^-19 / 6.626x10^-34

= 4.59x10¹⁴/sec

B) λ_o =c/v_o = 3x10⁸ / 4.59x10¹⁴ = 6.54x10^-7m

C) K.E of ejected electron = h(v-v_o) = hc(1/λ – 1/ λ_o)

=(6.626x3x10^-26) (1/500x10^-9 – 1/654x10^-9)

=(6.626x3x10^-26) / 10^-9(154/500x654)

= 9.36x10^-20J

k.E = 1mv²/2 = 9.36x10^-20J

=9.1x10^-31/2 = 9.36x10^-20J

Or

v2 = 20.55x10¹⁰m²s^-2

Or

v = 4.53x10⁵ms^-1

(a) Let us suppose threshold wavelength to be λ_D nm (= λ₀ x 10^-9 m), the kinetic energy of the radiation is given as:

h(v-v₀)=1/2 mv²

Or

hc(1/λ - 1/λ_D) = 1/2 mv²

Threshold wavelength (λ_D)= 540 nm

 Substituting this value in equation iii,we get

= h  x (3x10⁸)/10^-9 [1/400-1/540]

= [(9.11x10^-31)(5.20x10⁶)²] / 2

=6.66x10^-34Js

From the principle of conservation of energy, the energy of an incident photon (E) is equal to the sum of the work function (W₀) of radiation and its kinetic energy (K.E) i.e.,

Energy of incident radiation(E)= hc/λ = (6.626x10^-34)(3x10⁸)/(256.7x 10^-9)

= 7.74x10^-19 J

Since the potential applied gives the kinetic energy to the radiation ,therefore K.E of the electron = 0.35Ev

Therefore work function = 4.83 – 0.35 = 4.48 eV

Energy of incident photon (E) is given by,

= 10.2480 × 10^–17 J

= 1.025 × 10^–16 J

Energy with which the electron was bound to the nucleus

= 13.25 × 10^–16 J – 1.025 × 10^–16 J

= 12.225 × 10^–16 J

 = 12.225 × 10^–16 J/1.602 x10^-19 eV

= 7.63x10³eV

V = c/λ = 3x10⁸/1285x10^-9

= 3.29 x 10¹⁵(1/3² – 1/n²)

= 1/n2 = 1/9 - 3x10⁸/1285x10^-9 x (1 / 3.29 x 10¹⁵)

= 0.111- 0.071 = 0.04 or 1/25

Or

n² = 25,

n= 5

The spectrum lies in the infra-red region.

The radius of the n^th orbit of hydrogen-like particles = 0.529n²/Z Å

Now r₁ = 1.3225 nm or 1322.5 pm = 52.9n₁²

And

r₂ = 211.6pm = 52.9n₂²/Z

Therefore r₁/r₂ = 1322.5 / 211.6 = n₁²/n₂²

or n₁²/n₂² = 6.25

or n₁/n₂ = 2.5

therefore n₂ = 2 , n₁ = 5.

Thus the transition is from 5^th orbit to 2^nd orbit. It belongs Balmer series

Wave number  for the transition is given by,

1.097 × 10⁷ m^–1 (1/2²-1/5²)

=1.097 x 10⁷m^-1 (21/100)

= 2.303 × 10⁶ m^–1

Wavelength (λ) associated with the emission transition is given by,

= 0.434 ×10^–6 m

λ = 434 nm

the region is visible region

From de Broglie’s equation,

λ = h/mv

Substituting the values in this equation,we get

6.626x 10^-34 / (9.11x10^-31)(1.6x10⁶)

= 4.55 x 10^-10m

= 455pm

Mass of neutron = 1.675 x10^-27 kg

We know λ = h/mv

or

v = h/mλ

Substituting the value ,we get

6.626x10^-34 / (1.675x10^-27)(800x10^-12)

= 4.94x10⁴ m/sec

According to de Broglie’s equation,

λ = h/mv

Where,

λ = wavelength associated with the electron

h = Planck’s constant

m = mass of electron

v = velocity of electron

Substituting the values in the expression of λ:

λ = 6.626x10^-34 / (9.11x10^-31)(2.19x10⁶)

=3.32x10^-10

= 332 pm

Planck’s quantum theory = Based on the assumption that all atoms on the surface of the heated solid vibrate at the frequency, Planck developed a model that came to be known as Planck’s equation. Through experiments of frequencies and temperature, Planck was able to generate a constant, Planck’s constant

                                                           h = 6.62607 x 10^-34 J s 

Using this constant he was able to restate his theory: energy was directly proportional to frequency. He wrote his equation as 

                                                          E=hν

where E is energy, h is Planck’s constant, and v is frequency.

(i) Energy (E) of a photon is given by the expression,

E=hν

Where, h = Planck’s constant = 6.626 × 10^–34 Js

ν = frequency of light = 3 × 10¹⁵ Hz

Substituting the values in the given expression E = hv we get

E = (6.626 × 10^–34) (3 × 10¹⁵)

E = 1.988 × 10^–18 J

(ii) Energy (E) of a photon having wavelength (λ) is given by the expression,

 where , h = Planck’s constant = 6.626 × 10^–34 Js

c = velocity of light in vacuum = 3 × 10⁸ m/s

Substituting the values in the given expression of E:

λ = h/mv

= 6.626x10^-34 / (0.1)(4.37x10⁵)

= 1.515x10^-28m

From Heisenberg’s uncertainty principle,

Where,

Δx = uncertainty in position of the electron

Δp = uncertainty in momentum of the electron

Δx = 0.002nm = 2x10^-12m(given)

 Therefore, Substituting the values in the expression of Δp:

Δp = h/4π Δx or

= 2.637 × 10^–23 Jsm^–1

Δp = 2.637 × 10^–23 kgms^–1 (1 J = 1 kgms²s^–1)

Actual momentum = h/4πx 0.05nm

= 6.626x10^-34/4 x3.14 x 5 x10^-11

= 1.055 x 10^-24 kg m/sec

Quantum number – they are the index numbers which gives the complete address of an electron

Here n =principal quantum number

l = azimuthal quantum number

ms = Spin quantum number

For n = 4 and l = 2, the orbital occupied is 4d.

For n = 3 and l = 2, the orbital occupied is 3d.

For n = 4 and l = 1, the orbital occupied is 4p.

Hence, the six electrons i.e., 1, 2, 3, 4, 5, and 6 are present in the 4d, 3d, 4p, 3d, 3p, and 4p orbitals respectively.

Therefore, the increasing order of energies is 5(3p) < 2(3d) = 4(3d) < 3(4p) = 6(4p) < 1 (4d).

4p electrons, being farthest from the nucleus,experience the lowest effective nuclear charge.

Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of a multi-electron atom. The closer the orbital, the greater is the nuclear charge experienced by the electron (s) in it.

(i) 2s is closer to the nucleus than 3s.Hence 2s will experience larger effective nuclear charge.

(ii) 4d will experience greater nuclear charge than 4f since 4d is closer to the nucleus than 4f.

(iii) 3p will experience greater nuclear charge since it is closer to the nucleus than 3f because 3p is closer to nucleus than 3f.

Nuclear charge is defined as the net positive charge experienced by an electron in a multi-electron atom.

Silicon has greater nuclear charge(+14) than aluminium (+13).

Hence the unpaired 3p electron in case of silicon will experience more effective nuclear charge.

(a) Phosphorus (P): 1s2 2s2 2p6 3s2 3p3

No of unpaired electron = 3

(b) Silicon (Si): 1s2 2s2 2p6 3s2 3p2

No of unpaired electron = 2 (since p orbital can have maximum 6 electron )

(c) Chromium (Cr): 1s2 2s2 2p6 3s2 3p6 4s1 3d5

No of unpaired electrons = 6 (since 1 electron is to be added to 4s & 5 electron to be added to 3d orbital.

(d) Iron (Fe): 1s2 2s2 2p6 3s2 3p6 4s2 3d6

No of unpaired electrons = 4

(e) Krypton (Kr): 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6

Since all orbitals are fully occupied, there are no unpaired electrons in krypton.

(a) n = 4 (Given)

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

∴ l = 0, 1, 2, 3

Thus, four sub-shells are associated with n = 4, which are s, p, d and f.

(b) Number of orbitals in the n^th shell = n²

For n = 4

Number of orbitals = 16

Each orbital has one electron with m_s = -1/2

Hence there will be 16 electrons with m_s = -1/2

Given period = 2.0 x 10^-10s

We know, Frequency (ν) of light  ,therefore putting the values .we get

We also know

Where,c = velocity of light in vacuum = 3×10⁸ m/s

Substituting the value in the given expression of λ, we get

Wave number of light

Energy (E) of a photon = hc/λ

Where,λ = wavelength of light = 4000 pm = 4000 ×10^–12 m = 4x 10^-9 m

c = velocity of light in vacuum = 3 × 10⁸ m/s

h = Planck’s constant = 6.626 × 10^–34 Js

thefore the energy of photon (E)

= 6.626 × 10^–34 Js X 3 × 10⁸ m/s / 4x 10^-9 m = 4.965 x 10^-16 J

Now 4.965 x 10^-16 J is the energy of = 1 photon

Therefore 1 J will be the energy of = 1/4.965 x 10^-16 J = 2.014 x 10¹⁵ photons

(1)We know λ = 4 × 10^–7 m(given)

C = 3 x 108

From the equation E= hv or hc/ λ

Where, h = Planck’s constant = 6.626 × 10^–34 Js

c = velocity of light in vacuum = 3 × 10⁸ m/s

λ = wavelength of photon = 4 × 10^–7 m

Substituting the values in the given expression of E:

Hence, the energy of the photon is 4.97 × 10^–19 J.

(ii) The kinetic energy of emission Ek is given by

= (3.1020 – 2.13) eV

= 0.9720 eV

Hence, the kinetic energy of emission is 0.97 eV.

(iii) The velocity of a photoelectron (ν) can be calculated by the expression,

Where, (hv-hv₀) is the kinetic energy of emission in Joules and ‘m’ is the mass of the photoelectron. Substituting the values in the given expression of v:

v = 5.84 × 10⁵ ms^–1

Hence, the velocity of the photoelectron is 5.84 × 10⁵ ms^–1.