Here, \(T=30 \ mins.=0.5h, \ t=1.5hr. \ \therefore \ n=\frac { t }{ T } =\frac { 1.5 }{ 0.5 } =3\)
When distance is changed, count rate after 3 half lives is \(5{ s }^{ -1 }.\) Therefore, count rate at the beginning of 1.5 hr. \(={ \left( 2 \right) }^{ 3 }\times 5=40{ s }^{ -1 },\) which is \(\frac { 1 }{ 9 } \times 360{ s }^{ -1 }=\frac { 1 }{ 9 } th\) of original count rate. i.e. Intensity has been reduced by a factor of 9. As intensity \(\propto \frac { 1 }{ { \left( dist \right) }^{ 2 } } ,\)  therefore, distance of source from counter must have increased by a factor of 3 i.e. New distance \(=3\times 2=6m\)

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:

As, we know that
\(power=\frac { energy }{ time } =300\times { 10 }^{ 6 }W=3\times { 10 }^{ 8 }J/s\)
\( 170 \ MeV \ =170\times { 10 }^{ 6 }\times 1.6\times { 10 }^{ -19 }=27.2\times { 10 }^{ -12 }J\)
Number of atoms fissioned per second
\(=\frac { 3\times { 10 }^{ 8 } }{ 2.72\times { 10 }^{ -12 } } =\frac { 3\times { 10 }^{ 20 } }{ 27.2 } \)
Number of atoms fissioned per hour
\(=\frac { 3\times { 10 }^{ 20 }\times 3600 }{ 2.72 } =\frac { 3\times 36 }{ 27.2 } \times { 10 }^{ 22 }=4\times { 10 }^{ 22 }m\)

(i) Bohr's second postulate states that the electron revolves around the nucleus in certain privileged orbit which satisfy certain quantum condition that angular momentum of an electron is an integral multiple of  \(\cfrac { h }{ 2\pi } \), where h is Planck's constant. i.e  \(L=mvr=\cfrac { nh }{ 2\pi } \)
Where, m = mass of electron, v = speed of electron and r = radius of orbit of electron.
\(\therefore\) Circumference of electron in n^th orbit
= n \(\times\)de-Broglie wavelength associated with electron.  \(\left[ \because \lambda =\cfrac { h }{ mv } \right] \)            
(ii) Given, the electron in H-atom is initially in third excited state.
n = 4
And the total number of spectral lines of an atom that can exist is given by the relation
 \(=\ \cfrac { n(n-1) }{ 2 } \)  
Here,  n = 4
So, number of spectral lines = \(\cfrac { 4(4-1) }{ 2 } =\cfrac { 4\times 3 }{ 2 } =6\)
Hence, when a H-atom moves from third excited state to ground state, it emits six spectral lines. 

(i) The proton separation energy is given by
\({ S }_{ P }(sn)=({ M }_{ 119,70 }+{ M }_{ H }-{ M }_{ 120,70 }){ c }^{ 2 }\)
\(=(118.902199+1.0078252-119.902199){ C }^{ 2 }\)
\(=0.0114262{ C }^{ 2 }\)
\(Similarly,{ S }_{ p }(sb)=({ M }_{ 120,70 }+{ M }_{ H }-{ M }_{ 121,70 }){ c }^{ 2 }\)
\(=(119.902199+1.0078252-120.903822){ C }^{ 2 }\)
\(=0.006202 \ { c }^{ 2 }\)
\(Since,{ S }_{ p }(sn)>{ S }_{ p }(sb), \) Sn nucleus is more stable than sb nucleus.
(ii) The existence of magic number indicates that the shell structure of nucleus is similar to the shell structure of an atom. This also explains the peaks in binding energy/nucleon curve.