The square of the amplitude of the electron wave i.e \(\Psi ^{ 2 }\) at any point gives probability of finding an electron at that point.since, the region around the nucleus which represents the electron  density at different points is called an orbits, hence the wave function for an electron in an atom is called orbital wave function.

For g-orbitals, l = 4.
As for any value ‘n’ of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n – 1).
∴For l = 4, minimum value of n = 5

Energy (E) of one photon is given by E = hv
h = 6.626 x 10^-34 Js.
v = 5 x 10¹⁴ s^-1 (given)
E = (6.626 x 10^-34 Js) x (5 x 10¹⁴s^-1) = 3.313 x 10^-19 J
Energy of one mole of photons
= (3.313 x 10^-19 J) x (6.022 x 10²³mol^-1) = 199.51 KJ mol^-1

|ψ|² shows the probability of finding the electron at a point with coordinates (x, y, z).

Helium nucleus contains 2 protons and charge of a proton is same as that of an electron.
Therefore, the charge on the nucleus of a helium atom is (+2)×4.8×10⁻¹⁰=+9.6×10⁻¹⁰ esu

Heisenberg's uncertainty principle and de-Broglie's concept of dual nature of matter are two discoveris that put strong challenge to the Bohr model.

The line spectrum of any element has lines corresponding to definite wavelengths. Lines are obtained as a result of electronic transitions between the energy levels. Hence, the electrons in these levels have fixed energy, i.e quantised values.

In case of positive rays, the ions remaining after the loss of electrons might have the same magnitude of charge but different masses. Hence they will have different charge to mass ratio. Cathode rays are made up of electrons and all electrons have same charge to mass ratio. That's why charge to mass ratio of all cathode rays is same.

a) 4d = 4 + 2 = 6, 4f = 4 + 3 = 7, 5s = 5 + 0 = 5, 7p = 7 + 1 = 8
Hence, 5s has the lowest energy.
b) 5p = 5 + 1 = 6, 5d = 5 + 2 = 7, 5f = 5 + 3, 6s = 6 + 0 = 6, 6p = 6  + 1 = 7
Hence, 5f has highest energy.

(i) \({ 1s }^{ 2 }2s^{ 2 },{ 2p }_{ x }^{ 2 },{ 2p }_{ y }^{ 2 },{ 2p }_{ z }^{ 2 },3s^{ 2 }\)
(ii) \({ 1s }^{ 2 }2s^{ 2 },{ 2p }_{ x }^{ 1 },{ 2p }_{ y }^{ 1 },{ 2p }_{ z }^{ 1 }\)
(iii) \({ 1s }^{ 2 }2s^{ 2 },{ 2p }^{ 6 },3s^{ 2 },{ 3p }^{ 6 },4s^{ 2 },3d^{ 3 }\)
(iv) \({ 1s }^{ 2 }2s^{ 2 },{ 2p }^{ 6 },3s^{ 2 },{ 3p }^{ 6 },3d^{ 5 },4s^{ 1 }\)

Given, m = 100g = 0.1kg
v = 100km/h = \(\frac { 100\times 1000 }{ 60\times 60 } =\frac { 1000 }{ 36 } { ms }^{ -1 }\)   
From de-Broglie equation, wavelength,
\(\lambda =\frac { h }{ mv } =\frac { 6.626\times { 10 }^{ -34 }{ kgm }^{ 2 }{ s }^{ -1 } }{ 0.1kg\times \frac { 100 }{ 36 } { ms }^{ -1 } } =238.5\times { 10 }^{ -36 }m\)
As the wavelength is very small so wave nature cannot be detected.

We know that, mass of electron = 9.11 x 10^-31 kg
h = 6.626 x 10^-34 Js
Wavelength, 
\(\lambda =\frac { h }{ mv } =\frac { 6.626\times{ 10 }^{ -34 }kg \ { m }^{ 2 }{ s }^{ -1 } }{ 9.11\times{ 10 }^{ -31 }kg\times2.19\times{ 10 }^{ 6 }\times m{ s }^{ -1 } }\)
\( \lambda =3.32\times{ 10 }^{ -10 }m=332pm\)

Given, velocity of beam of helium atoms = 2.0\(\times \)10³m sec^-1
Mass of helium atom = \(\frac { 4 }{ 6.022\times { 10 }^{ 23 } } \)
= \(6.64\times { 10 }^{ -24 }g=6.64\times { 10 }^{ -27 }kg\)
According to de-Broglie equation, \(\lambda =\frac { h }{ mv } \)
\(=\frac { 6.626\times { 10 }^{ -34 }kg{ m }^{ 2 }{ s }^{ -1 } }{ (6.64\times { 10 }^{ -27 }kg)\times (2.0\times { 10 }^{ 3 }{ ms }^{ -1 }) } \)
\(= 4.99\times { 10 }^{ -11 }m=49.9pm \ [2]\)

Quantum numbers:-
Quantum numbers are basically a set of numbers that define the position and energy of an electron in an atom.
We know that:

According to the question, the electron is in 3d orbital.
This happens in the case where n=3.
And at n=3, value of l is 2 and the value of ml as we can see from the table lies within (-2to2) that is (-2,-1,0,1,2)
Therefore, the possible value of l is 2, n is 3 and m_l is = -2, -1, 0 +1, +2