2. n 2 = 3 to n 1 = 2

The transition in the hydrogen spectrum that would have the same wavelength as Balmer transition FROM n = 4 to n = 2 of He + spectrum is -

Chemistry - Structure Atom - Quiz-9

The number of electrons in the species $H_{2}^{+}, H_{2}, O_{2}^{+}$ , are respectively

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15, 2, 1
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15, 1, 2
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1, 2, 15
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2, 1, 15

Orbital that does not exist:

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6p
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2s
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3f
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2p

The set of quantum numbers which represent 3p is

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n =1, l =0;
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2, n = 3; l=1
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n = 4; l =2;
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n = 4; l = 3

The velocity associated with a proton moving at a potential difference of 1000 V is 4.37 × 10⁵ms^–1. If the hockey ball of mass 0.1 kg is moving with this velocity, then the wavelength associated with this velocity would be

$1 . 1.52 \times 10^{- 38} m 2 . 2.54 \times 10^{- 32} m 3 . 1.52 \times 10^{- 36} m 4 . 3.19 \times 10^{- 34} m$

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1
0%
2
0%
3
0%
4

The maximum wavelength in the Lyman series of He⁺ ion is -

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3R
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$\frac{1}{3 R}$
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$\frac{1}{R}$
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2R

According to the Bohr Theory, which of the following transitions in the hydrogen atom will
give rise to the least energetic photon?

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n = 5 to n = 3
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n = 6 to n = 1
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n = 5 to n = 4
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n = 6 to n = 5

The orbital angular momentum of a p-electron is given as -

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$\sqrt{3} \frac{h}{2 π}$
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$\sqrt{\frac{3}{2}} \frac{h}{π}$
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$\sqrt{6} \sqrt{\frac{h}{2 π}}$
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$\frac{h}{\sqrt{2} π}$

A 0.66 kg ball is moving with a speed of 100 m/s. The associated wavelength will be

$(h = 6.6 \times 10^{- 34} J s)$

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$6.6 \times 10^{- 34} m$
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$1.0 \times 10^{- 35} m$
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$1.0 \times 10^{- 32} m$
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$6.6 \times 10^{- 32} m$

The energy value is E = 3.03 × 10^–19 Joules, (h=6.6×10^–34 J x sec., C=3×10⁸ m/sec). The value of the corresponding wavelength is

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65.3 nm
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6.53 nm
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3.4 nm
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653 nm

Which of the following molecule is not paramagnetic :

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$C u^{+ +}$
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$F e^{2 +}$
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$C l^{-}$
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None of the above

The radius of hydrogen shell is 0.53Å, then in first excited state, radius of the shell will be :

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2.12 Å
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1.06 Å
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8.5 Å
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4.24 Å

Uncertainity in position of a $e^{-}$ and He is similar. If uncertainity in momentum of $e^{-}$ is $32 \times 10^{5}$ , then uncertainity in momentum of He will be :

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$32 \times 10^{5}$
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$16 \times 10^{5}$
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$8 \times 10^{5}$
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None of the above

In Hydrogen atom, energy of first excited state is – 3.4 eV. Then find out KE of same orbit of Hydrogen atom : -

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+ 3.4 eV
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+ 6.8 eV
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– 13.6 eV
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+ 13.6 eV

The value of Planck's constant is 6.63 × 10^–34Js. The velocity of light is 3.0 × 10⁸ ms^–1. The closest value to the wavelength in nanometers of a quantum of light with a frequency of $8 \times 10^{15} s^{- 1}$ is-

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$2 \times 10^{- 25}$
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$5 \times 10^{- 18}$
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$4 \times 10^{1}$
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$3 \times 10^{7}$

The mass and charge of one mole of electrons are, respectively,

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5.48 × $10^{- 7}$ kg, 9.65 C,
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5.48 × $10^{- 7}$ C, 1.098 x 10²⁷
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5.48 × $10^{- 7}$ kg, 9.65 × $10^{4}$ C
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9.65 × $10^{4}$ C, 1.098 x 10²⁷

The period in the modern periodic table indicates the value of :

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Atomic number
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Atomic mass
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Principal quantum number
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Azimuthal quantum number.

Number of orbitals indicated by following set of quantum, numbers, n = 3, l = 2, m =+2 is-

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1
0%

2
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3
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4

The maximum number of neutrons is present in the nuclei of

$1 . O_{8}^{16} 2 . {Mg}_{12}^{24} 3 . {Fe}_{26}^{56} 4 . {Sr}_{38}^{88}$

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1
0%
2
0%
3
0%
4

The complete symbol for the atom with the given atomic number (Z) and atomic mass (A) would be, respectively,
i. Z = 17 , A = 35.
ii. Z = 92 , A = 233.
iii. Z = 4 , A = 9.

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${}_{4}^{9}{Be}$ ${}_{92}^{233}U$ ${}_{4}^{9}{Be}$
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${}_{17}^{35}{Cl}$ ${}_{92}^{233}U$ ${}_{4}^{9}{Be}$
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${}_{4}^{9}{Be}$ ${}_{17}^{35}{Cl}$ ${}_{92}^{233}U$
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${}_{92}^{233}U$ ${}_{4}^{9}{Be}$ ${}_{17}^{35}{Cl}$

Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm.

The frequency (ν) and wave number of this yellow light would be, respectively,

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517 x 10¹⁴ s^-1 , 172 x 10⁶ m^-1
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6.17 x 10¹⁴ s^-1 , 1.72 x 10⁶ m^-1
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4.17 x 10¹⁴ s^-1 , 2.72 x 10⁶ m^-1
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5.17 x 10¹⁴ s^-1 , 1.72 x 10⁶ m^-1

The energy of the photon having frequency 3×10¹⁵ Hz would be

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1.988 × $10^{- 18}$ J
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1.588 × $10^{- 18}$ J
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1.288 × $10^{- 18}$ J
0%
2.988 × $10^{- 18}$ J

A photon of wavelength 4 × 10^–7 m strikes a metal surface, the work function of the metal being 2.13 eV. The kinetic energy of emission would be

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0.97 eV
0%
97 eV
0%
4.97 × $10^{- 19}$ eV
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5.84 × 10⁵ eV

The correct comparison between energy required to ionize an H atom if the electron occupies n = 5 orbits and the ionization enthalpy of H atom ( energy required to remove the electron from n =1 orbit) would be -

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More energy is required to ionize an electron in the 5th orbital of hydrogen atom as compared to that in the ground state
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More energy is required to ionize an electron in the 8th orbital of hydrogen atom as compared to that in the ground state
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Less energy is required to ionize an electron in the 5th orbital of hydrogen atom as compared to that in the ground state
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Less energy is required to ionize a neutron in the 6th orbital of hydrogen atom as compared to that in the ground state

The wavelength of the light emitted when the electron returns to the ground state in the H atom, from n = 5 to n = 1, would be: (The ground-state electron energy is –2.18 × 10^-18 ergs)

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Wavelength = 9.498 x 10^-8 km
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Wavelength = 12.498 x 10^-8 m
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Wavelength = 9.498 x 10^-8 m
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Wavelength = 9.498 x 10^-8 cm

The possible values of n, l, and m for the electron present in 3d would be respectively

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n = 3, l = 1, m = – 2, – 1, 3, 1, 2
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n = 3, l = 3, m = – 2, – 1, 0, 1, 2
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n = 3, l = 2, m = – 2, – 1, 0, 1, 2
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n = 5, l = 2, m = – 2, – 1, 0, 1, 2

Kinetic Energy of an electron is 3.0×10^–25 J. The wavelength of the electron will be-

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886.7 nm
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896.7 nm
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990.7 nm
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880.7 nm

Which of the following sets of quantum numbers is possible-

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n = 0, l = 0, m_l = 0, m_s = + ½
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n = 1, l = 0, m_l = 0, m_s = – ½
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n = 1, l = 1, m_l = 0, m_s = + ½
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n = 3, l = 3, m_l = –3, m_s = + ½

The total number of electrons in an atom with the following quantum numbers would be
(a) n = 4, m_s = – ½ (b) n = 3, l = 0

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16, 2
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11, 8
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16, 8
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12, 7

2. $n_{2}$ = 3 to n₁ = 2

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$n_{1} = 3 to n_{2} = 4$
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4. $n_{2}$ = 2 to n₁ = 1
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$n_{2}$ = 3 to n₁ = 1
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The transition in the hydrogen spectrum that would have the same wavelength as Balmer transition FROM n = 4 to n = 2 of He⁺ spectrum is -

2 ×10⁸ atoms of carbon are arranged side by side. The radius of a carbon atom if the length of this arrangement is 2.4 cm would be

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7.0 x 10^-11 m
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5.0 x 10^-11 m
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8.0 x 10^-11 m
0%

6.0 x 10^-11 m

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Practice Chemistry Quiz Questions and Answers

0:0:1

Practice Chemistry Quiz Questions and Answers

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