\(1 \ eV=1.602\times { 10 }^{ -19 }joule\)
\( 1.602\times { 10 }^{ -19 } \ joule=1 \ eV\)
\(1 \ joule=\frac { 1 }{ 1.602\times { 10 }^{ -19 } } eV\)
\(1 \ joule=6.242\times { 10 }^{ 18 }eV\)
\((ii) \ From \ E={ mc }^{ 2 }\)
\(m=\frac { E }{ { c }^{ 2 } } \therefore 1\ a.m.u.=\frac { 931\times 1.602\times { 10 }^{ -13 } }{ { \left( 3\times { 10 }^{ 8 } \right) }^{ 2 } }\)
\(=1.66\times { 10 }^{ -27 }kg\)
Now, definition, Mass of \(\ _{ 6 }{ { C }^{ 12 } }=12a.m.u.\)
\(=12\times 1.657\times { 10 }^{ -27 }\ kg=1.99\times { 10 }^{ -26 }kg\)

6 x \({ 10 }^{ 26 }\) Na atoms weights 23 kg.
Volume of target = \({ (10 }^{ -4 }\times{ 10 }^{ -3 })={ 10 }^{ -7 }\) \({ m }^{ 3 }\)
Density of sodium = (d) = 0.97 kg/\({ m }^{ 3 }\)
Volume of \({ 6\times10 }^{ 26 }\) Na atoms = \(\frac { 23 }{ 0.97 } { m }^{ 3 }=23.7{ m }^{ 3 }\)
Volume of occupied of 1 Na atom = \(\frac { 23 }{ 0.97\times6\times{ 10 }^{ 26 } } { m }^{ 3 }\)
\(=3.95\times{ 10 }^{ -26 }=2.53\times{ 10 }^{ 18 }\)
No of sodium atoms in the target 
 \(=\frac { { 10 }^{ -7 } }{ 3.95\times{ 10 }^{ -26 } } =2.53\times{ 10 }^{ 18 }\)
Number of photons/in the beam for \({ 10 }^{ -4 }{ m }^{ 2 }=n\) Energy per second
\(=nhv={ 10 }^{ -4 }J\times100={ 10 }^{ -2 }W\)
\(hv(for\lambda =660nm)=\frac { 1234.5 }{ 600 }\)
\(=2.05eV=2.05\times1.6\times{ 10 }^{ -19 }\)
\(\\ =3.28\times{ 10 }^{ -19 }J\)
\(n\frac { { 10 }^{ -2 } }{ 3.28\times{ 10 }^{ -19 } } =3.05\times{ 10 }^{ 16/s }\)
\( n=\frac { 1 }{ 3.2 } \times{ 10 }^{ 17 }=3.1\times{ 10 }^{ 16 }\)
If p is the probability of emission per atom , per photon,the number of photoelectrons emitted/second
\(=p\times3.1\times{ 10 }^{ 16 }\times2.53\times{ 10 }^{ 18 }\)
Current \(= p\times3.1\times{ 10 }^{ +16 }\times2.53\times{ 10 }^{ 18 }\times1.6\times{ 10 }^{ -19 }A\)
This must equal \(100\mu A\)
\(p=\frac { 100\times{ 10 }^{ -6 } }{ 1.25\times{ 10 }^{ +16 } } \)
\(\therefore \)  \(p=8\times{ 10 }^{ -21 }\)

Accoring to Bohr's theory, centripetal force required by the electron for its motion around the nucleus = Electric force between the proton and electron
\(\frac { mv^{ 2 } }{ r } =\frac { 1 }{ 4\pi \varepsilon _{ 0 } } .\frac { ({ q }_{ p })({ q }_{ e }) }{ { r }^{ 2 } } \) [From Coulomb's law]
where, r-atomic radius,q_p = charge of proton = +e
q_e = charge of electron = -e
\(\frac { mv^{ 2 } }{ r } =\frac { 1 }{ 4\pi \varepsilon _{ 0 } } .\frac { (e)(-e) }{ { r }^{ 2 } } =\frac { 1 }{ 4\pi \varepsilon _{ 0 } } .\frac { -e^{ 2 } }{ { r }^{ 2 } } \)
Now, given charge on proton, \({ q }_{ p }=+\frac { 4 }{ 3 } e\)
Charge on electron \({ q }_{ e }=-\frac { 3 }{ 4 } e\)
putting the new value (keeping after factors unchanged),
\(\frac { mv^{ 2 } }{ r } =\frac { 1 }{ 4\pi \varepsilon _{ 0 } } .\frac { \left( \frac { 4 }{ 3 } e \right) \left( -\frac { 3 }{ 4 } e \right) }{ { r }^{ 2 } } =\frac { 1 }{ 4\pi \varepsilon _{ 0 } } .\frac { -e^{ 2 } }{ { r }^{ 2 } } \)
i.e. Bohr's formula remain unchanged.

Let v₁ and v₂ be the velocities of the two particles of masses m₁ and m₂ respectively
According to law of conservation of linear momentum \(\ { m }_{ 1 }\overset { \rightarrow }{ { v }_{ 1 } } +{ m }_{ 2 }\overset { \rightarrow }{ { v }_{ 2 } } =M\times 0=0 \)
\( { m }_{ 1 }\overset { \rightarrow }{ { v }_{ 1 } } =-{ m }_{ 2 }\overset { \rightarrow }{ { v }_{ 2 } } \ or \ \left| { m }_{ 1 }\overset { \rightarrow }{ { v }_{ 1 } } \right| =\left| { m }_{ 2 }\overset { \rightarrow }{ { v }_{ 2 } } \right| \)
\( { \lambda }_{ 1 }=\frac { h }{ { m }_{ 1 }{ v }_{ 1 } } and{ \lambda }_{ 2 }=\frac { h }{ { m }_{ 2 }{ v }_{ 2 } } \)
\(or \ \frac { { \lambda }_{ 1 } }{ { \lambda }_{ 2 } } =\frac { { m }_{ 2 }{ v }_{ 2 } }{ { m }_{ 1 }{ v }_{ 1 } } =1\)

Ionisation energy is the minimum energy required to knock out an electron from an atom.Its value will be different for different atoms. Ionisation energy will also depend on the orbit from which electron is to be removed.
When an electron in hydrogen atom is replaced by a particle 200 times heavier than electron but having the same charge, ionisation energy will not change, as it depends only on charge and not on mass of particle.

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

\(6.25\times { 10 }^{ 11 }{ s }^{ -1 },0,5.0eV\)

\(Given \ h=6.6\times { 10 }^{ -34 }Js \ and \ c=3\times { 10 }^{ 8 }{ ms }^{ -1 }\)
2.486 eV

\(0.53\mathring { A } ;2.19\times { 10 }^{ 6 }{ ms }^{ -1 };6.6\times { 10 }^{ 15 }Hz\)