Explore topic-wise InterviewSolutions in Current Affairs.

(1) Dual character of the electromagnetic radiations i.e. wave like and particle like properties, and

(2) Atomic spectra explained only by assuming quantized electronic energy levels in atoms.

Correct option is  A) atoms

Correct option is  B) perpendicular to

A moving cricket ball will have a wave like character but the wave length of the wave would be very small. So, this concept cannot be applied to a moving cricket ball.

Number of waves in n^th orbit

\(\mathrm{\frac{Circumference\,of\,the\,orbit}{Wavelength}}\)

=2πr/λ = 2πr /(ℎ/mv) = 2πmvr/h

= 2π(nh /2π )/h =n

K.E. = Ze²/2r

P.E. = -Ze²/r

∴ P.E. = –2K.E

∴ K.E/P.E = - 1/2

Option : (C) \(-\frac{2.18\times 10^{-18}}{n^2}J\)

Wavelength of the radiation, λ = 6800 Å = 6800 x 10^-10 m

Work function of the metal, W₀ = hv₀ = hc/λ

= {6.626 x 10^-34 Js x 3 x 10⁸ ms^-1}/{6800 x 10^-10 m}

= 2.91 x 10^-19 J

and threshold frequency, v₀ = W₀/h = {2.92 x 10^-19 J}/{6.626 x 10^-34 Js} = 4.41 x 10¹⁴ s^-1

or, v₀ = 4.41 x 10¹⁴ Hz

Correct option is  A) Na⁺

Correct option is  B) N > O > C

∆E = E_∞ − E₁ 

= 0 − (−1312 kj mol⁻¹ ) 

= 1312 kj mol⁻¹

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

Mass of electron = 9.1 x 10^-31 kg = 9.1 x 10^-28 g

∴ 9.1 x 10^-28 g = 1 electron

or, 1 g = {1 electron x 1 g}/{9.1 x 10^-28 g} 

= 1.099 x 10²⁷ electron

Wavelength of radiation, λ = 242 nm = 242 x 10^-9 m

Energy (per mole) of the photons = {N_A.hc}/{λ}

= {6.023 x 10²³ mol^-1 x 6.626 x 10^-24 Js x 3 x 10⁸ ms^-1}/{242 x 10^-9 m}

= 4.95 x 10⁵ J mol^-1

= 494 kJ mol^-1

This energy is equal to the ionization energy of sodium. Therefore, Ionisation energy of sodium = 495 kJ mol^-1

E = hv = \(\frac{hc}{\lambda}\)

Wavelength \(\lambda\) = 400 nm = 400 × 10⁻⁹m

\(\therefore\) E = \(\frac{6.626\times10^{-34}Js\times3.0\times10^{-9}m}{400\times10^{-9}m}\)

= 4.969 × 10⁻¹⁹ J

Energy of one mole of photons

= 4.969 × 10⁻¹⁹ J x 6.022 × 10²³ mol⁻¹

= 29.923 × 10⁴

= 2.99 × 10⁵ J mol⁻¹

(i) The minimum energy needed to remove one mole of electrons from sodium

= (2.99 − 1.05) × 10⁵ J mol⁻¹

= 1.94 × 10⁵ J mol ^{−1​​​​​​​}

^{\(\therefore\)} The minimum energy for one electron from sodium

= \(\frac{1.94\times10^5Jmol^{-1}}{6.022\times10^{23}electron\,mol^{-1}}\)

= 0.322 × 10⁻¹⁸

= 3.22 × 10⁻¹⁹J

(ii) Maximum wavelength of light

\(\lambda=\frac{hc}{E}\)

= \(\frac{6.626\times10^{-34}Js\times3.0\times10^8ms^{-1}}{3.22\times10^{-19}J}\)

= 6.17 × 10⁻¹⁷m

= 617 × 10⁻⁹m

= 617 nm

According to Bohr's theory, Energy of electron in n^th orbital of H-atom with Atomic number (Z) = \(-13.6\frac{2^2}{n^2}\)eV

\(\therefore\) The energy of ground state electron in H-atom = -13.6 eV

Now, it absorbs 12.1 eV energy.

So, its energy increase to -13.6 + 12.1 = - 0.85 eV

\(\therefore\) E = -13.6\(\frac{z^2}{n^2}\)eV

-0.85 = -13.6 x \(\frac{(1)^2}{n^2}\) (\(\because\) for hydrogen Z = 1)

\(\therefore\) n² = \(\frac{13.6}{0.85}\) = 16

\(\therefore\) n = 4

Hence, the electron in H-atom jumps to 4th orbit on absorbing 12.1 eV energy.

Number of spectral lines = {n(n - 1)}/{2}

= {6(6 - 1)}/{2} = 30/2

= 15 emission lines.

Distance of 3^rd orbit

d₃ = \(\frac{52.9\times n^2}{z}\)

= \(\frac{52.9\times(3)^2}{1}\)

= 476.1 pm

Distance of 2^nd orbit

d₂ = \(\frac{52.9\times n^2}{z}\)

= \(\frac{52.9\times(2)^2}{1}\)

= 211.6 pm

\(\therefore\) Distance of separation between 3^rd and 4^th orbit of H-atom

= Distance of 3^rd orbit – Distance of 2^nd orbit from the nucleus

= 476.1 − 211.6

= 264.5 pm

Energy of the first orbit in a H atom = -2.18 x 10^-18 J atom^-1

Energy of the 5th orbit in a H atom = -{2.17 x 10^-18 J atom^-1}/{(5)²}

= -8.72 x 10^-10 J atom^-1

Correct option is  C) 10^-12

Photo electric effect: It is defined as the emission of electrons when light is shone onto a material. Electrons emitted in this manner are called photo-electrons. Photo electric effect on the basis of quantum theory:

(i) Electrons can only absorb or emit energy in discrete amounts called quanta (packets).

(ii) Energy of each quanta (photon) is directly proportional to frequency of radiation.

E= hv

Where E = Energy of photon

h = Planck′s constant

v = Frequency of radiation

Given, 

m = 60 gm = \(\frac{60}{1000}\)= 6 x 10^-2 kg

v = 10 m/sec

According to de-Broglie equation,

\(\lambda=\frac{h}{mv}\)

= \(\frac{6.626\times10^{-34}Js}{6\times10^{-2}kg\times10\,m/sec}\)

= 1.104 × 10⁻³³ m

C) Pauli’s exclusion principle

Name of the element is Aluminium (Al). its Valency is 3.

Noble gases having closed valence shell configuration as ns² np⁶ . The valence shell orbitais of noble gases are completely filled by electrons and it is very difficult to alter this stable arrangement by the addition or removal of electrons. Because of these reasons noble gases do not show much of chemical reactivity

The noble gases are placed in a separate group because these are unreactive and present in a very low concentration in the atmosphere and their properties do not resemble with any other group so they are placed in a separate group.

Correct option is  D) 107°

By sharing four electrons with other atoms.

B) 1s² 2s¹ 2p¹_x2p¹_y2p¹_z 

(i) Na, K, Li etc. are very reactive so these metals react vigorously with atmospheric oxygen to form oxides. Storing them in oil prevents their oxidation. 

(ii) Platinum, gold and silver are least reactive, so they are not corroded by air and water easily. 

(iii) Copper vessels get tarnished by reacting with air and water, due to which a layer is formed mainly contains black copper oxide and green copper hydroxide which is basic in nature. The citric acid in lemon dissolves this layer and washes off easily.

(a) A is sodium and B is chlo rine

(b) C is sodium chloride and D is sodium hydroxide

(c) C will have no effect on litmus solution since it is neutral in nature.

(d) B is a gas at room temperature.

(e) EB 3

Formaldehyde

(a) Atomic number .

(b) Periods.

(c) Alkali metals .

(d) Halogens .

(e) Noble Gases .

(f) Eighth.

Toluene is aromatic compound.

Common name:

formaldehyde

IUPAC name: methanal

 (d) both bromine and carboxylic group

Alcohol group;

Methanol

(b) They differ from other by CH₃ group.

(c) Environment

1. Antoine Lavoisier 

2. 60 

3. isomer 

4. Allotropy 

5. 7

The upper and lower homologue of C₂H₅OH are respectively propyl alcohol and methyl alcohol.

Ethanoic acid; CH₃COOH.

(a) Combustion

(b) Soap

(c) Detergent

(d) Ethanoic acid

The statement is-

False

Ethanol (C₂H₅OH)

(d) propyl alcohol and methyl alcohol

The vessels blacken due to deposits of black carbon particles on it which is caused due to incomplete combustion of fuel. Air holes are adjusted so that air enters through the holes and helps in complete combustion of the fuel.

Acetic acid has very low melting point i.e. 290 K, hence it freezes during winters in cold countries. So it is called glacial acetic acid.

Carbon forms large number of compounds because of tetravalency and catenation property. 

Tetravalency – Carbon has valency 4, to attain noble gas configuration carbon share its valence electrons with other elements like hydrogen, chlorine, etc. 

Catenation – Carbon also shows the property of self-linking in which it forms long, branched or cyclic chains to form large number of compounds.