In He atom two electrons of charge (-2e) revolve around the nucleus.
And \(E_n=-{me^2Z^2\over8\epsilon_0^2n^2h^2}=4E\ \ \ [Z=2^2=4]\)
Ground state energy
= 4 x E = 4 x (-13.6) = -54.4eV

Given, Bohr's radius, \({ r }_{ 1 }=5.3\times 10^{ -11 }\)
We know that, \({ r }_{ n }=n^{ 2 }{ r }_{ 1 }\)
Let  be radius of the orbit for n = 2
Therefore, \(r_{ 2 }=(2)^{ 2 }\times 5.3\times 10^{ -11 }\quad =2.12\times 10^{ -10 }m\)
(ii) Given, total energy of an electron in second excited state,
E = 1.51 eV
(a) Kinetic energy of electron is equal to negative of the total energy
\(\Rightarrow \)  K = -E = -(-1.51) = 1.51 eV
(b) Potential energy of electron is equal to negative of twice of its kinetic energy
\(\Rightarrow \) U = -2K = -2 x 1.51 = -3.02 eV

By considering the situation given in the question,
At t = 0, N_A = N₀ (maximum), while N_B = 0. As time increases, N_A decreases exponentially and the number of atoms of B increases. They become (N_B) maximum and finally drop to zero exponentially by radioactive decay law. So, graph showing the variation of number of atoms of A and B will be shown as below:

The energy of the gaseous hydrogen at room temperature are as given below:
E₁ = -13.6eV, E = -3.4eV
E₃ = -1.51eV, E₄  = - 0.85eV
E₃ - E₁ = - 1.51 - ( - 13.6) = 12.09eV
and     E₄ - E₁ = - 0.85 - ( - 13.6) = 12.75eV
As, both the values do not match the given value, but it is nearest to E₄ - E₁. Upto E₄ - E₁ energy level, the H-atoms would be excited. 
Lyman series, \(\cfrac { 1 }{ \lambda } =R\left[ \cfrac { 1 }{ { 1 }^{ 2 } } -\cfrac { 1 }{ { n }^{ 2 } } \right] \) 
For first member, n = 2
\(\therefore \)   \(\cfrac { 1 }{ { \lambda }_{ 1 } } =R\left[ \cfrac { 1 }{ { 1 }^{ 2 } } -\cfrac { 1 }{ { 2 }^{ 2 } } \right] =1.097\times { 10 }\left[ \cfrac { 4-1 }{ 4 } \right] \)
\(\Rightarrow \)  \({ \lambda }_{ 1 }=1.215\times { 10 }^{ -7 }m\)   
Balmer series, \(\cfrac { 1 }{ \lambda } =R\left[ \cfrac { 1 }{ { 2 }^{ 2 } } -\cfrac { 1 }{ { n }^{ 2 } } \right] \)                        
For first member, n = 3
\(\therefore \)  \(\cfrac { 1 }{ { \lambda }_{ 1 } } =R\left[ \cfrac { 1 }{ { 2 }^{ 2 } } -\cfrac { 1 }{ { 3 }^{ 2 } } \right] =1.097\times { 10 }\left[ \cfrac { 1 }{ 4 } -\cfrac { 1 }{ 9 } \right] \)
\(\Rightarrow \)  \({ \lambda }_{ 1 }=6.56\times { 10 }^{ -7 }m\)

The binding energy is H-atom,
\(E=\frac { { me }^{ 4 } }{ { \pi \varepsilon }_{ 0 }^{ 2 }{ h }^{ 2 } } =13.6 \ eV\quad ........(i)\)
If proton and neutron had charge e^' each and were governed by the same electrostatic force, then in the above equation we would need to replace electronic mass m by the reduced mass m^'  of proton-neutron and the electronic charge e by e^' .
\({ m }^{ ' }=\frac { M\times N }{ M+N } =\frac { M }{ 2 } \)
\( =\frac { 1836 \ m }{ 2 } =918 \ m\)
Here, M represents mass of a neutron/proton.
\(\therefore\) Binding energy \(=\frac { 918m{ \left( { e }^{ ' } \right) }^{ 4 } }{ { 8\varepsilon }_{ 0 }^{ 2 }{ h }^{ 2 } } =2.2\) MeV
Dividing Eqs. (ii) and (i), we get
\(918{ \left( \frac { { e }^{ ' } }{ e } \right) }^{ 4 }=\frac { 2.2 \ MeV }{ 13.6 \ eV }\)
\(=\frac { 2.2\times { 10 }^{ 6 } }{ 13.6 } \)
\( { \left( \frac { { e }^{ ' } }{ e } \right) }^{ 4 }=\frac { 2.2\times { 10 }^{ 6 } }{ 13.6\times 918 } =176.21\)
\( \left( \frac { { e }^{ ' } }{ e } \right) ={ \left( 176.21 \right) }^{ 1/4 }\)
\( =3.64\)