\({ 9 }^{ { F }^{ 19 } }={ 1s }^{ 2 }{ 2s }^{ 2 }{ 2p }_{ x }^{ 2 }{ 2p }_{ y }^{ 2 }{ 2p }_{ z }^{ 1 }\)
\({ 16 }^{ { S }^{ 32 } }={ 1s }^{ 2 }{ 2s }^{ 2 }{ 2p }^{ 6 }{ 3s }^{ 2 }{ 3p }_{ x }^{ 2 }{ 3p }_{ y }^{ 1 }{ 3p }_{ z }^{ 1 }\)
\( { 18 }^{ { Ar }^{ 38 } }={ 1s }^{ 2 }{ 2s }^{ 2 }{ 2p }^{ 6 }{ 3s }^{ 2 }{ 3p }^{ 6 }\)
Maximum number of unpaired electrons = 2 in\({ 16 }^{ { S }^{ 32 } }\)

a) [He]2s¹ represents ₃Li (lithium)
b) [Ne] 3s²3p³ represents ₁₅P (phosphorus)
c) [Ar]4s²3d¹ represent ₂₁Sc (scandium)

Given velocity of proton = 4.37 x 10⁵ m sec^-1
potential difference = 1000v
Mass of hockey ball = 0.1 kg
Wavelength associated with the velocity of hockey ball 
\(\lambda =\frac { h }{ mv } \)
\(= \frac { 6.626\times 10^{ -34 }\ kg\ m^{ 2 }\ s^{ -1 } }{ 0.1kg\times 4.37\times 10^{ 5 }m\ s^{ -1 } }\)
\(=15.16\times10^{ -39 } \ m=1.56\times10^{ -38 }\ m\)

\( \lambda=\frac{h}{m v} \)
\( \text { or } v=\frac{h}{m \lambda} \)
\( =\frac{6.63 \times 10^{-34} \mathrm{kgm}^2 \mathrm{~s}^{-1}}{\left(9.1 \times 10^{-31} \mathrm{~kg}\right) \times\left(100 \times 10^{-10} \mathrm{~m}\right)} \)
\( =7.28 \times 10^4 \mathrm{~ms}^{-1} . \)

E_n= \(\frac { \left( -2.18\times10^{ -18 }J \right) Z^{ 2 } }{ \left( n \right) ^{ 2 } } \) atom^-1
For He⁺, n = 1, Z = 2
 \(E_{ 1 }\frac { \left( -2.18\times10^{ -18 }J \right) Z^{ 2 } }{ \left( 1 \right) ^{ 2 } } =-8.72\times10^{ -18 }J\\ \)
The radius of the orbit is given by r_n = \(\frac{52.9 (n^2)}{Z} pm\)
r_n= \(\frac { \left( 0.0529nm \right) n^{ 2 } }{ Z } \)
Since, n = 1 and Z = 2
r_n= \(\frac { \left( 0.0529nm \right) 1^{ 2 } }{ Z } =0.02645 \ nm\)

1.67 x 10⁸ m^-1

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.

All radiations in vacuum travel with the same speed, i.e. 3 x 10⁸ ms^-1
Distance to be travelled from Mars to the Earth 
= 8 x 10⁷ km = 8 x 10⁷ x 10³ m(1 km = 10³m)
Time Taken = \(\frac { 8\times { 10 }^{ 7 }\times { 10 }^{ 3 } }{ 3\times { 10 }^{ 8 } } =2.66\times { 10 }^{ 2 }s=4min43s\)

Bohr's theory does not consider the de-Broglie concept of dual nature of electron and also contradicts with the Heisenberg's uncertaintly priciple , while the schrodinger equation is based on quantum mechanics while deals with the microscopic objects having both the particle as well as wave like character.

Ferquency, V = \(\frac { 1 }{ Period } \)= \(\frac { 1 }{ 2\ ns } \) = \(\frac { 1 }{ 2\times 10^{ -9 }s } \)
= 0.5 x 10⁹s^-1
Energy of the source = Energy of 1 photon x number of photons produced
E_source = hv x N
= 6.626 x 10^-34 Js x 0.5 x 10⁹ s^-1 x 2.5 x 10¹⁵
= 8.28 x 10^-10 J

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

No because velocity = 0 and thus, position can be measured accurately.

This is because according to Heisenberg, the position of a microscopic particle like electron cannot be determined with certainity.

This is because the energies of the orbits in which the electrons revolve are fixed.

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.

This means that the electrons in an atom have only definite values of energies.

As A = n + p
p = A - n = 13 - 7 = 6
Hence, atomic number, z = p = 6

Mass number is a whole number because it is the sum of the number of protons and number of neutrons whereas atomic mass is fractional because it is the average relative mass.

Neutron is a neutral practice. Hence it will not be deflected on passing through an electric field.

Cathode rays originate from the cathode whereas anode rays are not obtained from the anode. They are produced from the gaseous atoms by knock out of the electrone with high speed cathode rays.

To find the wavelength of the emitted light, we use the following formula:
\(\dfrac{1}{\lambda } = R\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]{Z^2}\)
Where λ is the wavelength, RH is the Rydberg constant, n1 and n2 are the lower and upper orbits involved in transition respectively and Z is the atomic number of the species.
For lowest frequency in the Lyman series in the hydrogen atom spectrum, n1 = 1 and n2 = 2. Substituting the rest of the values, \({R_H} = 1.09677 \times {10^7}{m^{ - 1}}\) and [Z = 1], we get:
\(\dfrac{1}{\lambda } = 1.09677\left[ {\dfrac{1}{{1_{}^2}} - \dfrac{1}{{{2^2}}}} \right]{1^2}\)
On simplifying, we get:
\(\dfrac{1}{\lambda } = 1.097 \times {10^7}\left( {\dfrac{3}{4}} \right) = 0.82275 \times {10^7}{m^{ - 1}}\)
Therefore, wavelength
\(\lambda = \dfrac{1}{{0.82275 \times {{10}^7}}} = 1.216 \times {10^{ - 7}}m\)
Simplifying further, we get:
\( \Rightarrow \lambda = 121.6 \times {10^{ - 9}}m = 121.6nm\)
As we know, \(\nu = \dfrac{c}{\lambda }\)
Where v is the frequency and x is the speed of light
\(\Rightarrow \nu = \dfrac{{3 \times {{10}^8}m{s^{ - 1}}}}{{121.6 \times {{10}^{ - 9}}m}}\)
On simplifying, we get:
ν=2.46 × 10¹⁵s⁻¹
For finding energy, we use the formula E=h/cλ
Where E is the energy and hℎ is the Planck’s constant
\(\Rightarrow E=\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{121.6 \times 10^{-9}}\)
On solving, we get \(E=16.3 \times 10^{-19} \mathrm{~J}\)
As we can see from our first equation, \(\frac{1}{\lambda} \propto Z^2\)
\(\Rightarrow \lambda \propto \frac{1}{Z^2}\)
Therefore, to get the wavelength of the spectral line of \(\mathrm{Li}^{2+}\) we just need to divide the wavelength of hydrogen atom with the square of the atomic number of Lithium (3)
\(\lambda_{L i}=\frac{121.6}{3^2} n m=13.55 n m\)
Therefore, energy of this transition:
\(E=\frac{h c}{\lambda_{L i}} \Rightarrow E=\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{13.55 \times 10^{-9}}\)
On solving, we get:
\(E=146.7 \times 10^{-19} \mathrm{~J}\)

(i) Total electrons if n = 4 = \(2n^{2}\),2 x \(4^{2}=32\)
half of the total electrons,i.e 16 electrons have ms = – ½
(ii) n = 3, l = 0 it is 3s-orbital and it can have two electron.