Physics Nuclei

Apply energy conservation,

$U_i+K_i=U_f+K_f$

$\Rightarrow -\frac{GMm}{R}+K_i=-\frac{GMm}{3R}+\frac{1}{2}m{v}^2$

$\Rightarrow -\frac{GMm}{R}+K_i=-\frac{GMm}{3R}+\frac{1}{2}\times m\times \frac{GM}{3R}$

$\Rightarrow K_i=-\frac{1}{6}\frac{GMm}{R}+\frac{GMm}{R}$

$K_i=\frac{5}{6}\frac{GMm}{R}$

$A=Ao{e}^{-\lambda t}$

$\Rightarrow 2250=4250e^{-\lambda t}$

$\Rightarrow \lambda \times 10\times 0.434=0.274$

$\Rightarrow \lambda =\frac{0.274}{10\times 0.434}=0.63mi{n}^{-1}$

$A\to 286$

$Z=81$

Nuclear force is related to quarks and gluons in the following ways:

• Protons and neutrons are made up of a combination of smaller particles called quarks and down quarks. A proton has two up quarks and one down quark whereas neutron has one up quark and two down quarks.
• Nuclear force is an interaction

Nuclear force is important to maintain the stability of an atomic nucleus for the following reasons:

• A nucleus has positively charged sub-particles called protons. Since like charges repel each other, a nucleus, in ideal scenario, should fall apart because protons won't be able to coexist in the nucl

As we all know that like charges repel each other. This means that there must be a repulsive electromagnetic force between positively charged protons. However, the nuclear force is said to be 100 times stronger than electromagnetic force which makes it strong enough to hold protons together within a

$A=-\frac{dN}{dt}=\lambda N=\lambda N_0{e}^{-\lambda t}\Rightarrow \lambda \frac{N_0}{16}=\lambda N_0{e}^{-80\lambda }\Rightarrow \frac{1}{16}=e^{-80\lambda }\Rightarrow \frac{1}{16}=e^{-80\lambda }$

$\Rightarrow 80\lambda =\mathcal{l}n\left(16\right)=4\mathcal{l}n\left(2\right)\Rightarrow t_{\frac{1}{2}}=\frac{\mathcal{l}n\left(2\right)}{\lambda }=\frac{80}{4}=20days$

As we know that equivalent half life of radioactive material decays by simultaneous emissions, is given as

$\frac{1}{T}=\frac{1}{T_1}+\frac{1}{T_2}=\frac{1}{700}+\frac{1}{1400}=\frac{3}{1400}\Rightarrow T=\frac{1400}{3}years$

$\Rightarrow t=\frac{\mathcal{l}n\left(3\right)}{\lambda }=\frac{\mathcal{l}n\left(3\right)}{\mathcal{l}n\left(2\right)}T=\frac{1.1}{0.693}\times \frac{1400}{3}\approx 740.74years$

Using conservation linear momentum, we can write

$Mv=\frac{h}{\lambda }=\frac{h\nu }{c}\Rightarrow v=\frac{h\nu }{Mc}$                                                 

$R=-\frac{dN}{dt}=\lambda N=\lambda N_0{e}^{-\lambda t}\Rightarrow \mathcal{l}n\left(R\right)=\mathcal{l}n\left(\lambda N_0\right)-\lambda t,so$

$\lambda =tan\theta =\frac{6}{40}=\frac{3}{20}{s}^{-1}\Rightarrow$ Slope of graph

$\Rightarrow t_{\frac{1}{2}}=\frac{0.693}{\lambda }=\frac{0.693\times 20}{3}=4.62s$