To write the electronic configuration of an atom, it is better if we remember the atomic number of noble gases and the orbitals follow the noble gas. The atomic number of $$Ca$$  is 20 and its nearest noble gas is argon $$(Ar = 18).$$
Hence, the electronic configuration of $$Ca = \left[ {Ar} \right]4{s^2}.$$

$$\eqalign{ & {\left( {K.E.} \right)_1} = h{\upsilon _1} - h{\upsilon _0} \cr & {\left( {K.E.} \right)_2} = h{\upsilon _2} - h{\upsilon _0} \cr & {\text{As}}\,\,\,{\left( {K.E.} \right)_1} = 2 \times {\left( {K.E.} \right)_2} \cr & \therefore \,\,\left( {h{\upsilon _1} - h{\upsilon _0}} \right) = 2\left( {h{\upsilon _2} - h{\upsilon _0}} \right) \cr} $$
$${\text{or}}$$  $${\upsilon _0} = 2{\upsilon _2} - {\upsilon _1} = 2 \times \left( {2 \times {{10}^{16}}} \right) - \left( {3.2 \times {{10}^{16}}} \right)$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.8 \times {10^{16}}\,Hz\,\,\,{\text{or}}\,\,\,8 \times {10^{15}}\,Hz$$

$$\eqalign{ & {\text{Power of the bulb}} = 100\,watt = 100\,J\,{s^{ - 1}} \cr & {\text{Energy of one photon}}\,\,E = h\upsilon = \frac{{hc}}{\lambda } \cr & = \frac{{6.626 \times {{10}^{ - 34}}Js \times 3 \times {{10}^8}m\,{s^{ - 1}}}}{{400 \times {{10}^{ - 9}}\,m}} \cr & = 4.969 \times {10^{ - 19}}\,J \cr & {\text{Number of photons emitted}} \cr & = \frac{{100\,J\,{s^{ - 1}}}}{{4.969 \times {{10}^{ - 19}}J}} \cr & = 2.012 \times {10^{20}}{s^{ - 1}} \cr} $$

The energy of second Bohr orbit of hydrogen atom $$\left( {{E_2}} \right)$$  is $$ - 328\,kJ\,mo{l^{ - 1}}$$
$$\eqalign{ & \,\,\,\,\,\,\,{E_n} = - \frac{{1312}}{{{n^2}}}kJ\,mo{l^{ - 1}} \cr & \therefore \,\,{E_2} = - \frac{{1312}}{{{2^2}}}kJ\,mo{l^{ - 1}} \cr & {\text{If}}\,\,n = 4 \cr & \therefore \,\,{E_4} = - \frac{{1312}}{{{4^2}}}kJ\,mo{l^{ - 1}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - 82\,kJ\,mo{l^{ - 1}} \cr} $$

The radius of nucleus is of the order of $$1.5 \times {10^{ - 13}}$$   to $$6.5 \times {10^{ - 13}}cm$$    or 1.5 to 6.5 Fermi $$\left( {1\,{\text{Fermi}} = {{10}^{ - 13}}cm} \right)$$

$$\eqalign{ & \Delta E = 2.178 \times {10^{ - 18}}\left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) = \frac{{hc}}{\lambda } \cr & 2.17 \times {10^{ - 18}} \times \frac{3}{4} = \frac{{hc}}{\lambda } = \frac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{\lambda } \cr & \lambda = \frac{{6.62 \times \times {{10}^{ - 34}}}}{{2.17 \times {{10}^{ - 18}}}}\frac{{3 \times {{10}^8} \times 4}}{{ \times 3}} \cr & = 1.214 \times {10^{ - 7}}m \cr} $$

The fifth $$d$$ - orbital $$\left( {{d_{{z^2}}}} \right)$$  has a doughnut shaped electron cloud in the centre whereas others have clover leaf shape.

$$\eqalign{ & {\text{de Broglie wavelength}}\,\,\lambda = \frac{h}{{mv}} \cr & \frac{{{\lambda _1}}}{{{\lambda _2}}} = \frac{{{m_2}{v_2}}}{{{m_1}{v_1}}};\,\frac{1}{4} = \frac{1}{9} \times \frac{{{v_2}}}{{{v_1}}} \cr & \frac{{{v_2}}}{{{v_1}}} = \frac{9}{4};\frac{{{v_1}}}{{{v_2}}} = \frac{4}{9} \cr & KE = \frac{1}{2}m{v^2} \cr & \frac{{K{E_1}}}{{K{E_2}}} = \frac{{{m_1}}}{{{m_2}}} \times \frac{{v_1^2}}{{v_2^2}} \cr & = \frac{9}{1} \times {\left( {\frac{4}{9}} \right)^2} \cr & = \frac{{16}}{9} \cr} $$

According, to Pauli’s exclusion principle “no two electrons in an atom can have the same values of all the four quantum numbers.”
In $$1{s^2}$$
$$\eqalign{ & {\text{for }}I{\text{ electron}}\,\,n = 1,\,l = 0,m = 0,s = + \frac{1}{2} \cr & {\text{for }}II{\text{ electron}}\,\,n = 1,\,l = 0,m = 0,s = - \frac{1}{2} \cr} $$
It means if the values of $$n, l,$$  and $$m$$ are same, then the value of spin quantum number must be different, i.e. $$ + \frac{1}{2}$$  and $$ - \frac{1}{2}.$$

Shell with $$n = 3$$   has $$3s,3p\left( {{p_x},{p_y},{p_z}} \right)$$    and $$3d$$ $$\left( {{d_{xy}},{d_{xz}},{d_{yz}},{d_{{x^2} - {y^2}}},{\text{and}}\,\,{d_{{z^2}}}} \right)$$      orbitals.
$$s$$  has no nodal plane.
Each of $${{p_x},{p_y},{p_z}}$$  has one nodal plane, which means a total of three nodal planes.
$${d_{{z^3}}}$$  has no nodal plane.
Each of $${{d_{xy}},{d_{xz}},{d_{yz}},{d_{{x^2} - {y^2}}}}$$    has two nodal planes, which means a total of eight nodal planes.
Hence, for $$n = 3,$$  a total of 11 nodal planes are there.