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જવાબ : All have equal energy.
જવાબ : Similarities:
જવાબ : Electron has particle nature
જવાબ : The electronic configuration of the element palladium (Z = 46) is [Kr]36 4d10 5S°.
This means that it has no impaired electron.
જવાબ : A quantum is a bundle of energy of a definite magnitude (E = hv) and it may be from any source. However, a photon is quantum of energy associated with light only.
જવાબ : The energy of the electron is regarded as zero when it is at infinite distance from the nucleus.
At that point force of attraction between the electron and the nucleus is almost nil. Therefore, its energy is regarded as zero.
જવાબ : The metals with low ionisation enthalpies are used in photoelectric cells. Caesium (Cs), an alkali metal belonging to group 1 is the most commonly used metal.
જવાબ : I = 0,1, 2
m =- 2, — 1, 0, + 1, + 2 and S = +1/2 and-1/2
for each value of m.
જવાબ : Both have identical shape, consisting of four lobes.
જવાબ : Ground state means the lowest energy state. When the electrons absorb energy and jump to outer orbits, this state is called excited state.
જવાબ : Balmer series.
જવાબ : ‘l’ signifies the secondary quantum number.
‘L’ signifies second energy level (n = 2).
જવાબ : Orbitals having same energy belonging to the same subshell.
જવાબ : In the construction of electron microscope used for the measurement of objects of very small size.
જવાબ : As Fe3+ contains 5 impaired electrons while Fe2+ contains only 4 unpaired electrons. Fe3+ is more paramagnetic.
જવાબ : Hydrogen.
જવાબ : 6.62 x 1034 Js.
જવાબ : Cosmic rays > X-rays > radio waves.
જવાબ : Electron has wave nature.
જવાબ : Z = 35
A = 80
Atomic no. = 35 No. of protons = 35
No. of protons No. of electrons No. of neutrons
= 80 – 35 = 45
જવાબ : α < p < e.
જવાબ : 15
જવાબ : 18
જવાબ : I represents the subshell and L represents shell.
જવાબ : s-orbital
જવાબ : h/4π
જવાબ : Fe (Z = 26) : [Ar]18 3d64s2
Fe2+ion : [Ar]18 3d6
No. of unpaired electrons = 4
The relation is known as de Broglie’s relationship.
(ii) For hydrogen atom ; rn = 0.529 x n2 Å
r5 = 0.529 x (5)2 = 13.225 Å = 1.3225 nm.
જવાબ : According to uncertainty principle,
જવાબ : According to uncertainty principle,
જવાબ : (i)Boron (Z = 5) ; 1s2 2s2 1p1
(ii)Neon (Z = 10) ; 1s2 2s2 2p6
(iii)Aluminium (Z = 13) ; 1s2 2s2 2p6 3s2 3p1
(iv)Chlorine(Z = 17) ; 1s2 2s2 2p6 3s2 3p5
(v)Calcium (Z = 20) ; 1s2 2s2 2p6 3s2 3p6 4s2
(vi)Rubidium (Z = 37) ; 1s2 2s2 2p6 3s2 3p6 3d104s2 4p6 5s1.
જવાબ : (i) Mass of an electron = 9.1 × 10-28 g
9.1 × 10-28 g is the mass of = 1 electron
(ii) One mole of electrons = 6.022 × 1023 electrons
Mass of 1 electron = 9.1 × 10-31 kg
Mass of 6.022 × 1023 electrons = (9.1 × 10.31kg) × (6.022 × 1023) = 5.48 × 10-7 kg
Charge on one electron = 1.602 × 10-19 coulomb
Charge on one mole electrons = 1.602 × 10-19 × 6.022 × 1023 = 9.65 × 104 coulombs
જવાબ : (i) One mole of methane (CH4) has molecules = 6.022 × 1023
No. of electrons present in one molecule of CH4 = 6 + 4 = 10
No. of electrons present in 6.022 × 1023 molecules of CH4 = 6.022 × 1023 × 10
= 6.022 × 1024 electrons (ii) Step I. Calculation of total number of carbon atoms
Gram atomic mass of carbon (C-14) = 14 g = 14 × 103 mg
14 × 103 mg of carbon (C-14) have atoms = 6.022 × 1023
Step II. Calculation of total number and tatal mass of neutrons
No. of neutrons present in one atom (C-14) of carbon = 14 – 6 = 8
No. of neutrons present in 3-011 × 1020 atoms (C-14) of carbon = 3.011 × 1020 × 8
= 2.408 × 1021 neutrons
Mass of one neutron = 1.675 × 10-27 kg
Mass of 2.408 × 1021 neutrons = (1.675X10-27 kg) × 2.408 × 1021
= 4.033 × 10-6 kg. (iii) Step I. Calculation of total number ofNH3 molecules
Gram molecular mass of ammonia (NH3) = 17 g = 17 × 103 mg
17 × 103 mg of NH3 have molecules = 6.022 × 1023
Step II. Calculation of total number and mass of protons
No. of protons present in one molecule of NH3 = 7 + 3 = 10 . No. of protons present in 12.044 × 1020 molecules of NH3 = 12.044 × 1020 × 10
= 1.2044 × 1022 protons
Mass of one proton = 1.67 × 10-27 kg
Mass of 1.2044 × 1022 protons = (1.67 × 10-27 kg) × 1.2044 × 1022
= 2.01 × 10-5 kg.
No, the answer will not change upon changing the temperature and pressure because only the number of protons and mass of protons are involved.
જવાબ : (i) Energy of photon (E) = hv
h = 6.626 × 10-34 J s ; v = 3 × 1015 Hz = 3 × 1015s-1
∴ E = (6.626 × 10-34 J s) × (3 × 1015 s-1) = 1.986 × 1018 J
Energy of photon (E) = hv = hcλ
h = 6.626 × 10 34 J s; c = 3 × 108 m s-1 ;
λ= 0.50 Å = 0.5 × 10-10 m.
જવાબ : Energy of photon (E) = hcλ
h = 6.626 × 10-34 Js, c = 3 × 108 m s-1, λ = 4000 pm = 4000 × 10-12 = 4 × 10-9 m
જવાબ : The maximum no. of emission lines = n(n–1)2 = 6(6–1)2 =3 × 5 = 15
જવાબ : Cosmic rays < X-rays < amber colour < microwave < FM
જવાબ : Time duration (t) = 2 ns = 2 x 10-9 sEnergy of one photon, E = hv = (6.626 x 10-34 Js) x (109/2 s-1) = 3.25 x 10-25J
જવાબ : Let the no. of electrons in the ion = x
∴ the no. of the protons = x + 3 (as the ion has three units positive charge)
and the no. of neutrons = x + x×31.7100 = xc + 0.304 x
Now, mass no. of ion = No. of protons + No. of neutrons
= (x + 3) + (x + 0.304x)
∴ 56 = (x + 3) + (x + 0.304x) or 2.304x = 56 – 3 = 53
x = 532.304 = 23
Atomic no. of the ion (or element) = 23 + 3 = 26
The element with atomic number 26 is iron (Fe) and the corresponding ion is Fe3+.
જવાબ : Let the no. of electron in the ion = x
∴ the no. of protons = x – 1 (as the ion has one unit negative charge)
and the no. of neutrons = x + x×11.1100 = 1.111 x
Mass no. or mass of the ion = No. of protons + No. of neutrons
(x – 1 + 1.111 x)
Given mass of the ion = 37
∴ x- 1 + 1.111 x = 37 or 2.111 x = 37 + 1 = 38
x = 382.111 = 18
No. of electrons = 18 ; No. of protons = 18 – 1 = 17
Atomic no. of the ion = 17 ; Atom corresponding to ion = Cl
Symbol of the ion = 3717Cl–
જવાબ : An element can be identified by its atomic number only. Let us find the atomic number.
Let the number of protons = x
Number of neutrons = x + [x×31.7100 = (x × 0.317x)
Now, Mass no. of element = no. of protons =no. neutrons
81 = x + x + 0-317 x = 2.317 x or x = 812.317 = 35
∴ No. of protons = 35, No. of neutrons = 81 – 35 =46
Atomic number of element (Z) = No. of protons = 35
The element with atomic number (Z) 35 is bromine 8135Br
જવાબ : In the symbol BAX of an element :
A denotes the atomic number of the element
B denotes the mass number of the element.
The atomic number of the element can be identified from its symbol because no two elements can have the atomic number. However, the mass numbers have to be mentioned in order to identify the elements. Thus,
Symbols 7935Br and 79Br are accepted because atomic number of Br will remain 35 even if not mentioned. Symbol 3579Br is not accepted because atomic number of Br cannot be 79 (more than the mass number = 35). Similarly, symbol 35Br cannot be accepted because mass number has to be mentioned. This is needed to differentiate the isotopes of an element.
જવાબ : We have studied that in Rutherford’s experiment by using heavy metals like gold and platinum, a large number of a-particles sufferred deflection while a very few had to retrace their path.If a thin foil of lighter atoms like aluminium etc. be used in the Rutherford experiment, this means that the obstruction offered to the path of the fast moving a-particles will be comparatively quite less. As a result, the number of a-particles deflected will be quite less and the particles which are deflected back will be negligible
જવાબ : (a) Radius of zinc atom =Undefined control sequence \AA= 1.3 Å = 1.3 × 10-10m = 130 × 10-12m = 130 pm
(b) Length of the scale = 1.6 cm = 1.6 × 1010 pm
Diameter of zinc atom = 260 pm
જવાબ :For H atom (Z = 1), En =2.18 × 10-18 × (l)2 J atom-1 (given)
જવાબ : The length of the arrangement = 2.4 cm
Total number of carbon atoms present = 2 ×108
Thus, the circumference (2πr) of the Bohr orbit for hydrogen atom is an integral multiple of the de Broglie wavelength.
જવાબ : (a) For n = 4
Total number of electrons = 2n2 = 2 × 16 = 32
Half out of these will have ms = —1/2
∴ Total electrons with ms (-1/2) = 16
(b) For n = 3
l= 0 ; ml = 0, ms +1/2, -1/2 (two e–)
જવાબ : (i) The set of quantum numbers is not possible because the minimum value of n can be 1 and not zero.
(ii) The set of quantum numbers is possible.
(iii) The set of quantum numbers is not possible because, for n = 1, l can not be equal to 1. It can have 0 value.
(iv) The set of quantum numbers is not possible because for l = 0. mt cannot be + 1. It must be zero.
(v) The set of quantum numbers is not possible because, for n = 3, l ≠ 3.
(vi) The set of quantum numbers is possible.
જવાબ : (a) 1s orbital
(b) 4f orbital
(c) 3p orbital
(d) 4d orbital
જવાબ : (i) For n = 3; l = 0, 1 and 2.
For l = 0 ; ml = 0
For l = 1; ml = +1, 0, -1
For l = 2 ; ml = +2, +1,0, +1, + 2
(ii) For an electron in 3rd orbital ; n = 3; l = 2 ; ml can have any of the values -2, -1, 0,
+ 1, +2.
(iii) 1p and 3f orbitals are not possible.
જવાબ : (i) (a) 1s2
(b) 1s2 2s2 2p6
(ii) (a) Na (Z = 11) has outermost electronic configuration = 3s1
(b) N (Z = 7) has outermost electronic configuration = 2p3
(c) Fe (Z = 26) has outermost electronic configuration = 3d6
(iii) (a) Li
જવાબ : Na+ and Mg2+ are iso-electronic species (have 10 electrons) K+, Ca2+ , S2- are iso-electronic species (have 18 electrons)
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