Atoms and Nuclei

$ - 13.6 + 10.2 = {{ - 13.6} \over {{n^2}}}$

$ \Rightarrow {{13.6} \over {{n^2}}} = 3.4$

$ \Rightarrow n = 2$

$ \Rightarrow \Delta L = 2 \times {h \over {2\lambda }} - 1 \times {h \over {2\lambda }}$

$ = {h \over {2\lambda }}$

$ \Rightarrow \Delta L \simeq 1.05 \times {10^{ - 34}}$ Js

${{dA} \over {dt}} - ( - {\lambda _1}A) + ( - {\lambda _2}A)$

$ \Rightarrow {{dA} \over {dt}} = - ({\lambda _1} + {\lambda _2})A$

$ \Rightarrow {\lambda _{eff}} = {\lambda _1} + {\lambda _2}$

$ \Rightarrow {{\ln 2} \over {{{({t_{1/2}})}_{eff}}}} = {{\ln 2} \over {{{({t_{1/2}})}_1}}} + {{\ln 2} \over {{{({t_{1/2}})}_2}}}$

$ \Rightarrow {({t_{1/2}})_{eff}} = {{4.5 \times 3} \over {7.5}}$ hours = 1.8 hours

${}_{92}{U^{238}}\buildrel {} \over \longrightarrow {}_{82}P{b^{206}} + x\left( {{}_2H{e^4}} \right) + {}_y\left( {{}_{ - 1}{\beta ^0}} \right)$

$238 = 206 + 4x + 0$

$ \Rightarrow 4x = 32 \Rightarrow x = 8$

also, $92 = 82 + 2x - y$

$y = 82 + 16 - 92 = 6$

We know that

$R = {R_0}{A^{1/3}}$

$ \Rightarrow \underbrace {\ln \left( {{R \over {{R_0}}}} \right)}_y=\underbrace {{1 \over 3}}_m\underbrace {\ln (A)}_x$

$\Rightarrow$ Straight line

We know that $v \propto {Z \over n}$

$\Rightarrow$ Required ratio $ = {{{2 \over 3}} \over {{1 \over 3}}}$

$ = 2:1$

$T.E. = {{ - {Z^2}m{e^4}} \over {8{{\left( {nh{\varepsilon _0}} \right)}^2}}}$

$P.E. = {{ - {Z^2}m{e^4}} \over {4{{\left( {nh{\varepsilon _0}} \right)}^2}}}$

$K.E. = {{{Z^2}m{e^4}} \over {8{{\left( {nh{\varepsilon _0}} \right)}^2}}}$

As electron makes transition to higher level, total energy and potential energy increases (due to negative sign) while the kinetic energy reduces.

An atom based on classical theory of Rutherford's model should collapse as the electrons in continuous circular motion that is a continuously accelerated charge should emit EM waves and so should lose energy. These electrons losing energy should soon fall into heavy nucleus collapsing the whole atom.

²²⁰A $\to$ ¹⁰⁵B + ¹¹⁵C

$\Rightarrow$ Q = [105 $\times$ 6.4 + 115 $\times$ 6.4] $-$ [220 $\times$ 5.6] MeV

$\Rightarrow$ Q = 176 MeV