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જવાબ : All have equal energy.

જવાબ : **Similarities:**

- Both have spherical shape.
- Both have same angular momentum.

- 1s has no node while 2s has one node.
- Energy of 2s is greater than that of 1s.

જવાબ : Electron has particle nature

જવાબ : The electronic configuration of the element palladium (Z = 46) is [Kr]^{36 }4d^{10} 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.

If n is equal to 3, what are the values of quantum numbers l and m?

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જવાબ : I = 0,1, 2

m =- 2, — 1, 0, + 1, + 2 and S = +1/2 and-1/2

for each value of m.

What is common between d_{xy }and d_{x2-y 2 }orbitals?

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જવાબ : Both have identical shape, consisting of four lobes.

What is the difference between ground state and excited state?

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જવાબ : Ground state means the lowest energy state. When the electrons absorb energy and jump to outer orbits, this state is called excited state.

Which series of lines of the hydrogen spectrum lie in the visible region?

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જવાબ : Balmer series.

What is difference between the notations l and L?

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જવાબ : ‘l’ signifies the secondary quantum number.

‘L’ signifies second energy level (n = 2).

What are degenerate orbitals ?

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જવાબ : Orbitals having same energy belonging to the same subshell.

What is the most important application of de Broglie concept?

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જવાબ : In the construction of electron microscope used for the measurement of objects of very small size.

Which one Fe^{3+}, Fe^{2+}is more paramagnetic and why?

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જવાબ : As Fe^{3+} contains 5 impaired electrons while Fe^{2+} contains only 4 unpaired electrons. Fe^{3+} is more paramagnetic.

Which element does not have any neutron?

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જવાબ : Hydrogen.

What is value of Planck’s constant in S.I. units?

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જવાબ : 6.62 x 10^{34 }Js.

Arrange X-rays, cosmic rays and radio waves according to frequency.

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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} 3d^{6}4s^{2}

Fe^{2+}ion : [Ar]^{18} 3d^{6}

No. of unpaired electrons = 4

જવાબ :

The relation is known as de Broglie’s relationship.

જવાબ :

જવાબ :

**(ii)** For hydrogen atom ; r_{n} = 0.529 x n^{2} Å

r_{5} = 0.529 x (5)^{2} = 13.225 Å = 1.3225 nm.

જવાબ : According to uncertainty principle,

જવાબ : According to uncertainty principle,

જવાબ :

જવાબ : (i)Boron (Z = 5) ; 1s^{2} 2s^{2} 1p^{1}

(ii)Neon (Z = 10) ; 1s^{2} 2s^{2} 2p^{6}

(iii)Aluminium (Z = 13) ; 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{1}

(iv)Chlorine(Z = 17) ; 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{5}

(v)Calcium (Z = 20) ; 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2}

(vi)Rubidium (Z = 37) ; 1s^{2} 2s^{2} 2p^{6 }3s^{2 }3p^{6} 3d^{10}4s^{2} 4p^{6} 5s^{1}.

જવાબ :

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 × 10^{23} electrons

Mass of 1 electron = 9.1 × 10^{-31} kg

Mass of 6.022 × 10^{23} electrons = (9.1 × 10.31kg) × (6.022 × 10^{23}) = 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 × 10^{23} = 9.65 × 10^{4} coulombs

જવાબ : **(i)** One mole of methane (CH_{4}) has molecules = 6.022 × 10^{23}

No. of electrons present in one molecule of CH_{4} = 6 + 4 = 10

No. of electrons present in 6.022 × 10^{23} molecules of CH_{4} = 6.022 × 10^{23} × 10

= 6.022 × 10^{24} electrons
**(ii)** **Step I.** Calculation of total number of carbon atoms

Gram atomic mass of carbon (C-14) = 14 g = 14 × 10^{3} mg

14 × 10^{3} mg of carbon (C-14) have atoms = 6.022 × 10^{23}

**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 × 10^{20} atoms (C-14) of carbon = 3.011 × 10^{20} × 8

= 2.408 × 10^{21} neutrons

Mass of one neutron = 1.675 × 10^{-27} kg

Mass of 2.408 × 10^{21} neutrons = (1.675X10^{-27} kg) × 2.408 × 10^{21}

= 4.033 × 10^{-6} kg.
**(iii) Step I.** Calculation of total number ofNH_{3} molecules

Gram molecular mass of ammonia (NH_{3}) = 17 g = 17 × 10^{3} mg

17 × 10^{3} mg of NH_{3} have molecules = 6.022 × 10^{23}

**Step II.** Calculation of total number and mass of protons

No. of protons present in one molecule of NH_{3} = 7 + 3 = 10 .
No. of protons present in 12.044 × 10^{20} molecules of NH_{3} = 12.044 × 10^{20} × 10

= 1.2044 × 10^{22} protons

Mass of one proton = 1.67 × 10^{-27} kg

Mass of 1.2044 × 10^{22} protons = (1.67 × 10^{-27} kg) × 1.2044 × 10^{22}

= 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 :

જવાબ :

જવાબ :

જવાબ : **(i)** Energy of photon (E) = hv

h = 6.626 × 10^{-34} J s ; v = 3 × 10^{15} Hz = 3 × 10^{15}s^{-1}

∴ E = (6.626 × 10^{-34} J s) × (3 × 10^{15} s^{-1}) = 1.986 × 10^{18} J

Energy of photon (E) = hv = hcλ

h = 6.626 × 10 34 J s; c = 3 × 10^{8} m s^{-1} ;

λ= 0.50 Å = 0.5 × 10^{-10} m.

જવાબ :

જવાબ : Energy of photon (E) = hcλ

h = 6.626 × 10^{-34} Js, c = 3 × 10^{8} m s^{-1}, λ = 4000 pm = 4000 × 10^{-12} = 4 × 10^{-9} m

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

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 :

જવાબ :

જવાબ :

જવાબ : Time duration (t) = 2 ns = 2 x 10^{-9} s

No. of photons = 2.5 x 10

∴ Energy of source = 3.3125 x 10

જવાબ :

જવાબ :

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 Fe^{3+}.

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 ^{79}Br 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 negligibleજવાબ : **(a)** Radius of zinc atom =Undefined control sequence \AA= 1.3 Å = 1.3 × 10^{-10}m = 130 × 10^{-12}m = 130 pm

**(b)** Length of the scale = 1.6 cm = 1.6 × 10^{10} pm

Diameter of zinc atom = 260 pm

જવાબ :

જવાબ :

જવાબ :

For H atom (Z = 1), EFor He

જવાબ :

જવાબ : The length of the arrangement = 2.4 cm

Total number of carbon atoms present = 2 ×10^{8}

જવાબ :

જવાબ :

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 = 2n^{2} = 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 ; m_{l} = 0, m_{s} +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 ; m_{l} = 0

For l = 1; m_{l} = +1, 0, -1

For l = 2 ; m_{l} = +2, +1,0, +1, + 2

**(ii)** For an electron in 3rd orbital ; n = 3; l = 2 ; m_{l} can have any of the values -2, -1, 0,

+ 1, +2.

**(iii)** 1p and 3f orbitals are not possible.

જવાબ : **(i) (a)** 1s^{2}

**(b)** 1s^{2} 2s^{2} 2p^{6}

**(c)** 1s^{2}2s^{2}2p^{6}

**(d)** 1s^{2}2s^{2}2p^{6}.

**(ii) (a**) Na (Z = 11) has outermost electronic configuration = 3s^{1}

**(b)** N (Z = 7) has outermost electronic configuration = 2p^{3}

**(c)** Fe (Z = 26) has outermost electronic configuration = 3d^{6}

**(iii)** **(a)** Li

**(b)** P

**(c)** Sc

જવાબ : Na^{+} and Mg^{2+} are iso-electronic species (have 10 electrons) K^{+}, Ca^{2+} , S^{2-} are iso-electronic species (have 18 electrons)

જવાબ :

જવાબ :

જવાબ :

જવાબ :

-.

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- Math Book for CBSE Class 11
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- Chemistry II Book for CBSE Class 11
- Physics I Book for CBSE Class 11
- Physics II Book for CBSE Class 11
- Biology Book for CBSE Class 11

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