\$\alpha\$-particle consists of :

\$\alpha \$ - particle is nucleus of Helium which has two protons and two neutrons.Hence option A is correct answer.

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>>Class 12
>>Physics
>>Atoms
>>Alpha Particle & Rutherford Model
>> $Geiger Marsden experiment$

Geiger Marsden experiment

Q: Ionized hydrogen atoms and \$\alpha\$-particles with same momenta enters perpendicular to a constant magnetic field, B. The ratio of their radii of their paths \$r_H : r_{\alpha}\$ will be :

We know,\$r_H = \dfrac{P}{e\times B}\$\$r_{\alpha} = \dfrac{P}{2e\times B}\$(As \$\alpha\$ has 2e of charge on it)\$\dfrac{r_H}{r_{\alpha}} = \dfrac{\dfrac{P}{eB}}{\dfrac{P}{2eB}}\$\$\implies \dfrac{r_H}{r_{\alpha}} = \dfrac{2}{1}\$Hence option A is correct answer

Q: A proton and an alpha particle having the same kinetic energy are allowed to pass through a uniform magnetic field perpendicular to the direction of their motion. Compare the radii of the paths of proton and alpha particle.

a proton has a mass= m & charge = qmass of an alpha particlem' = 4m & charge q' = 2qK.e of proton = K.e of alpha particle\$0.5mv^2=0.5m'{v'}^2\$\$v^2=4{v'}^2\$\$\dfrac{v}{v'}=\sqrt{4}=2\$radius of charged particle in uniform magnetic field is \$r=\dfrac{mv}{qB}\$\$\dfrac{r}{r'}=\dfrac{m}{m'}\times \dfrac{v}{v'} \times \dfrac{q'}{q}=\dfrac{1}{4}\times \dfrac{2}{1} \times \dfrac{2q}{q}=\dfrac{1}{1}\$

Q: Determine the distance of the closest approach when an alpha particle of kinetic energy \$4.5 \, MeV\$ strikes a nucleus of \$Z = 80\$, stops, and reverses its direction.

Let r be the center to center distance between the alpha particle and the nucleus (\$Z = 80\$). When the alpha particle is at the stopping point, then\$K=\dfrac{1}{4\pi\epsilon_0} \,\dfrac{(Ze)(2e)}{r}\$or \$r=\dfrac{1}{4\pi\epsilon_0} \,\dfrac{(2Ze^2)}{K}\$\$=\dfrac{9 \times 10^9\times2\times80\,e^2}{4.5\,MeV}=\dfrac{9 \times 10^9\times2\times80\,(1.6\times10^{-19})^2}{4.5 \times 10^{6}\times1.6\times10^{-19}\,J}\$\$=\dfrac{9 \times 160 \times 1.6 }{4.5}\times10^{-16}=512 \times 10^{-16}m\$\$=5.14 \times 10^{-14}\,m\$

Q: When an \$\alpha\$-particle of mass \$'m'\$ moving with velocity \$'v'\$ bombards on a heavy nucleus of charge \$'Ze'\$, its distance of closest approach from the nucleus depends on \$m\$ as:

At closest distance of approach, the kinetic energy of the particle will completely convert into potential energy.\$\Rightarrow \dfrac{1}{2}mv^2=\dfrac{KQq}{d}\$\$\Rightarrow d\propto \dfrac{1}{m}\$

Q: An alpha nucleus of energy \$\displaystyle\frac{1}{2}mv^2\$ bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to

Let the closest distance upto which the alpha particle approaches be \$r\$.At this distance the kinetic energy of the particle is zero.Given: Initial kinetic energy of the particle, \$K.E_i = \dfrac{1}{2}mv^2\$Also initial potential energy of the system is assumed to be zero.Atomic number of alpha particle, \$Z_{\alpha} = 2\$Final potential energy of the system at closest distance, \$P.E_f = \dfrac{K (Ze) 2e}{r}\$Using conservation of energy : \$K.E_i + P.E_i = K.E_f +P.E_f\$\$\therefore\$ \$\dfrac{1}{2}mv^2 + 0 = 0 + \dfrac{K2Ze^2}{r}\$\$\implies\$ \$r = \dfrac{K 4Ze^2}{mv^2} \propto \dfrac{1}{m}\$ (where, \$K = \dfrac{1}{4\pi \epsilon_o}\$)

Q: The distance of closest approach of an \$\alpha\$- particle fired towards a nucleus with momentum \$p\$,is \$r\$. The distance of closest approach when the momentum of \$\alpha\$-particle is \$2p\$ is

\$\dfrac{1}{2}mv^{2}=\ \dfrac{q_{1}\ q_{2}}{4\pi \epsilon _{0\ r}}\$Initially, \$\ \dfrac{p^{2}}{2m}= \dfrac{2e\ q}{4\pi \epsilon _{0}\ r}\$ ----(1)Now, \$\ \dfrac{( 2p)^{2}}{2m} = \dfrac{2eq}{4\pi \epsilon _{0}\ r^{1}}\$ ----(II)Equation \$\ (I)\div (ii)\$\$\dfrac{1}{4} = \dfrac{r^{1}}{r}\$\$r^{1}=\dfrac{r}{4}\$

Q: An \$\alpha-\$ particle having energy \$10 MeV\$ collides with a nucleus of atomic number \$50\$.Then distance of closet approach will be

At the distance of closest approach the whole kinetic energy get converted into potential energy.So \$ KE=(1/4 \pi \epsilon_ 0)2Ze^2/r\$Where r is distance of closest approach.\$r=\dfrac{2Ze^2}{KE\times 4\pi \epsilon_0}\$putting \$Z=50, (1/4\pi \epsilon_0)=9\times10^9 ,e=1.6\times 10^{-19}, KE=10\times 1.6\times 10 ^{-13} Joule\$we get \$r=14.4\times 10^{-15}m\$Option D is correct.

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