The basic purpose of scattering experiment is defeated because solid hydrogen will be much lighter target compared to the alpha particle acting as a projectile. According to the theory of elastic collisions, the target hydrogen will move much faster compared to alpha, after the collision. We cannot determine the size of hydrogen nucleus.

\(Here; \ K.E.=\frac { 1 }{ 2 } { m\upsilon }^{ 2 }=0.0327 \ eV=0.0327\times 1.6\times { 10 }^{ -19 }J\)
\(\\ or \ \ \frac { 1 }{ 2 } \times \left( 1.675\times { 10 }^{ -27 } \right) \times { \upsilon }^{ 2 }=0.0327\times 1.6\times { 10 }^{ -19 }\)
\(On \ solving, \ we \ get \ \upsilon \ = \ 2.5\times { 10 }^{ 3 }m/s.\)
\(Time, \ t=\frac { distance }{ velocity } =\frac { 10 }{ 2.5\times { 10 }^{ 3 } } =4\times { 10 }^{ -3 }s\)
\(Now \ N={ N }_{ 0 }{ \left( \frac { 1 }{ 2 } \right) }^{ t/T }or \ \frac { N }{ { N }_{ 0 } } ={ \left( \frac { 1 }{ 2 } \right) }^{ t/T }={ \left( \frac { 1 }{ 2 } \right) }^{ \frac { 4\times { 10 }^{ -3 } }{ 700 } }=0.999952\)
\(\therefore \ Fraction \ of \ neutron \ decayed \ = \ 1-0.999952 \ = \ 0.000048\)

25.7 MeV

\(1.2\times { 10 }^{ 20 };\ 1.55\times { 10 }^{ 14 }disintegrations/sec\)

\(1.782\times { 10 }^{ 10 }years\)

400 days.

0.511 MeV,  941.1 MeV
\({ E }_{ 1 }={ m }_{ e }{ c }^{ 2 }=9.1\times { 10 }^{ -31 }{ \left( 3\times { 10 }^{ 8 } \right) }^{ 2 }J\)
\(=\frac { 81.9\times { 10 }^{ -15 } }{ 1.6\times { 10 }^{ -13 } } MeV=0.511 \ MeV\)
\({ E }_{ 2 }={ m }_{ p }{ c }^{ 2 }=1.673\times { 10 }^{ -27 }{ \left( 3\times { 10 }^{ 8 } \right) }^{ 2 }J\)
\(=\frac { 1.673\times { 9\times 10 }^{ -11 } }{ 1.6\times { 10 }^{ -13 } } MeV=941.1 \ MeV\)

\(3646.8\mathring { A } \)
\(Take \ { n }_{ 1 }=2 \ and \ { n }_{ 2 }=\infty \)

\(1.22\times { 10 }^{ 4 }V\)

According to Bohr's quantum condition, the permitted orbits are those in which the angular momentum of the electron is integral multiple of \(nh/2\pi\), where h is Planck's constant
i.e. m v r = \(nh/2\pi\) (when n = 1, 1, ............ called principal quantum number).

Balmer series of \(H_\gamma\) required transition from n=5 to n = 1
So energy required to excite electron fromn=1 to n = 5 is 
E = E_1E_5 = 13.6-0.54 = 13.06 eV
if angular momentum is conserved, then angular momentum of photon
= L₅-L₂ = 5\(\hbar\)-2\(\hbar\)= 3\(\hbar\)
= 3x1.06 x 10^-34kg m²s^-1
= 3.18 x 10^-34 kg m²s^-1

\(1.23\times { 10 }^{ 4 }{ s }^{ -1 }\)