Nuclear charge is defined as the net positive charge experienced by an electron in a multielectron atom. The higher the atomic number, the higher is the nuclear charge. Silicon has 14 protons while aluminium has 13 protons. Hence, silicon has a larger nuclear charge of (+14) than aluminium, which has a nuclear charge of (+13). Thus, the electrons in the 3p orbital of silicon will experience a more effective nuclear charge than aluminium.
\({ 13 }^{ Al }={ 1s }^{ 2 },{ 2s }^{ 2 },{ 2p }^{ 6 },{ 3s }^{ 2 },{ 3p }^{ 1 }\)
\({ 14 }^{ Si }={ 1s }^{ 2 },{ 2s }^{ 2 },{ 2p }^{ 6 },{ 3s }^{ 2 },{ 3p }^{ 2 }\)

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= \(\frac { \left( 0.0529nm \right) 1^{ 2 } }{ Z } =0.02645 \ nm\)

Frequency, \(v=\frac { c }{ \lambda } \)
\(\because \) 1 nm = 10^-9m
\(\therefore \) 580nm = 580 x 10^-9 m = 580 x 10^-7cm
\(v=\frac { 3.0\times 10^{ 8 }{ ms }^{ -1 } }{ 580\times { 10 }^{ -9 }m } =5.17\times { 10 }^{ 14 }{ s }^{ -1 }\)
\( [Velocity \ of \ light=3\times { 10 }^{ 8 }{ ms }^{ -1 }]\)
Wave number,
 \(\bar { v } =\frac { 1 }{ \lambda } =\frac { 1 }{ 580\times 10^{ -7 }cm } \)
\( =1.724\times { 10 }^{ 4 }cm^{ -1 }\)

(A) \(\lambda =300nm=300\times { 10 }^{ -9 }m\)
(B) \(\lambda =300\mu m=300\times { 10 }^{ -6 }m\)
(C) \(\lambda =3nm=3\times { 10 }^{ -9 }m\)
(D) \(\lambda =30\overset { \circ }{ A } =30\times { 10 }^{ -9 }m=3\times { 10 }^{ -9 }m\)
\(\because \) Energy, \(E=\frac { hc }{ \lambda } orE\propto \frac { 1 }{ \lambda } \) 
\(\therefore \) Increasing order of energy is B

\(\lambda =\frac { c }{ v } =\frac { 3.0\times { 10 }^{ 8 }ms^{ -1 } }{ 4.620\times { 10 }^{ 14 }{ s }^{ -1 } } =649.4nm\)
Thus, it lies in the visible light.