Number of unpaired electrons in N²⁺ is

2
Zero
1
3

Solution

N(7) = 1s² 2s² 2p³

N²⁺ = 1s², 2s² 2p¹_x

Unpaired electrons = 1.

Which one of the following represents noble gas configuration :

1s², 2s² 2p⁶, 3s² 3p⁶ 3d¹⁰, 4s² 4p⁶ 4d¹⁰, 5s² 5p⁶ 5d⁶, 6s²
1s², 2s² 2p⁶, 3s² 3p⁶ 3d¹⁰, 4s² 4p⁶ 4d¹⁰, 5s² 5p⁶ 5d¹, 6s²
1s², 2s² 2p⁶, 3s² 3p⁶ 3d¹⁰, 4s² 4p⁶ 4d¹⁰, 5s² 5p⁶
1s², 2s² 2p⁶, 3s² 3p⁶ 3d¹⁰, 4s² 4f¹⁴, 5s² 5p⁶ 5d¹

Solution

Noble gas must have 8 electrons in its outer most shell. Among the given configuration only option (c) has 8 electron in its outermost orbital. i.e. the configuration is 2, 8, 18, 18, 8. So option (c) is the right answer.

The wavelength of a moving electron having 4.55 × 10^-25 J of kinetic energy is :

7.27 × 10^-7 metre
72.7 × 10^-7 metre
7.27 × 10^-9 metre
72.7 × 10^-9 metre

Solution

When a metal surface is exposed to solar radiations

The emitted electrons have energy less than a maximum value of energy depending upon frequency of incident radiations
The emitted electrons have energy less than maximum value of energy depending upon intensity of incident radiation
The emitted electrons have zero energy
The emitted electrons have energy equal to energy of photons of incident light

The mass of a photon with a wavelength equal to 1.54 × 10^-8 cm is

0.8268 × 10^-34 kg
1.2876 × 10^-33 kg
1.4285 × 10^-32 kg
1.8884 × 10^-32 kg

Solution

The de-Broglie wavelength of an electron in the ground state of hydrogen atom is : [K.E. = 13.6 eV; 1eV = 1.602 × 10^-19J]

33.28 nm
3.328 nm
0.3328 nm
0.0332 nm

Solution

The ratio of the radius of the first Bohr orbit for the electron orbiting the hydrogen nucleus to that of the electron orbiting the deuterium nucleus(mass nearly twice that of the hydrogen nucleus) is approximately

2 : 1
1 : 1
1 : 2
4 : 1

Solution

If the radius of first Bohr orbit be a₀, then the radius of the third orbit would be

3 × a₀
6 × a₀
9 × a₀
1/9 × a₀

Solution

Radius of nth orbit r_n = ^a₀⁄_Z × n² ;

∴ r₃ = 9a₀

Which of the following is related with both wave nature and particle nature ?

Interference
E = mc²
Diffraction
E = hv

Solution

(i) Interference and diffraction support the wave nature of electron.

(ii) E = mc² supports the particle nature of electron.

(iii) E = hv = ^hc⁄_λ is de-Broglie equation and it supports both wave nature and particle nature of electron.

The de Broglie wavelength of a tennis ball of mass 60 g moving with a velocity of 10 metres per second is approximately

Planck’s constant, h = 6.63 × 10^–34Js

10^–31 metres
10^–16 metres
10^–25 metres
10^–33 metres

Solution

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