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Find thd ratio of distances of closest approach of a proton and an alpha particle projected towards the same nucleus with the same initial kinetic energy.


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Answer 1

⇝ To Find :-

The ratio of distances of closest approach of a proton and an alpha particle projected towards the same nucleus with the same initial kinetic energy.

⇝ Solution :-

We Know,

$\text m_ \alpha  = 4\text{amu}$

$\text m_\text p = 1\text{amu}$

❒ Calculating Kinetic Energies :

⏩ For α - particle :

$\text{K.E}_\alpha  =  \frac{1}{2} \text m_ \alpha\text  v_ \alpha  {}^{2}$

$\text{K.E}_\alpha  =  \frac{1}{2} \times 4\times \text  v_ \alpha  {}^{2}$

$\large\green{ \pmb{\bf{K.E}_\alpha  = 2 v_ \alpha  {}^{2}}}$

⏩ For proton :

$\text{K.E}_\text p  =  \frac{1}{2} \text m_ \text p\text  v_ \text p  {}^{2}$

$\text{K.E}_\text p  =  \frac{1}{2} \times 1\times \text  v_ \text p  {}^{2}$

$\large\green{ \pmb{\bf{K.E}_\text p  = \dfrac{1}{2} v_ \text p  {}^{2}}}$

As Given in Question kinetic energy is same of both α - particle and proton :

$\dfrac{\text K.\text E_ \alpha }{\text K.\text E_\text p}  = 1$

$:\longmapsto \frac{2 \times  {\text v_ \alpha }^{2} }{ \frac{1}{2} \times\text  v_\text p  {}^{2}  } = 1$

$:\longmapsto \text v_\text p {}^{2}  = 4 \text v_ \alpha  {}^{2}$

$\purple{ \Large :\longmapsto  \underline {\boxed{{\pmb{\bf v_p = 2v_\alpha}} }}}$

❒ We Have 3rd Equation of Motion as :

$\pink{ \bigstar \:  \: } \large \bf \orange{ \underbrace{ \underline{ {v}^{2} -  {u}^{2} = 2as  }}}$

⏩ Applying 3rd Equation For α - particle

$:\longmapsto\text v_ \alpha  {}^{2}  - \text u_ \alpha  {}^{2}  = 2\text a\text s_ \alpha$

As Initial Velocity is 0

$:\longmapsto\text v_ \alpha  {}^{2} - 0 = 2\text a\text s_ \alpha$

$\red{\large \pmb{ \bf 2as_ \alpha  = v_ \alpha  {}^{2} }} \:  -  -  - (1)$

⏩ Applying 3rd Equation For proton :

$:\longmapsto\text v_ p  {}^{2}  - \text u_ p  {}^{2}  = 2\text a\text s_ p$

As Initial Velocity is 0

$:\longmapsto\text v_ p  {}^{2} - 0 = 2\text a\text s_ p$

$\red{\large{ \bf 2as_ p  = v_ p  {}^{2} }} \:  -  -  - (2)$

⏩ Dividing (2) by (1) :

$\pmb{\bf\dfrac{s_{\alpha}}{s_p} =  \frac{ {v_ \alpha }^{2} }{v_p {}^{2} } }$

Putting $\pmb{\bf v_p = 2v_\alpha}$

$:\longmapsto \sf\dfrac{s_{\alpha}}{s_p} =  \frac{ {v_ \alpha }^{2} }{(2v_ \alpha)  {}^{2} }$

$:\longmapsto \sf\dfrac{s_{\alpha}}{s_p} =  \frac{{ \cancel{v_ \alpha }^{2}} }{4 \:  \cancel{v_ \alpha  {}^{2} }}$

$\purple{ \Large :\longmapsto  \underline {\boxed{{\pmb{\dfrac{s_{p}}{s_ \alpha } =  \frac{4}{1} } }}}}$

Hence,

$\large\underline{\pink{\underline{\frak{\pmb{Required  \: Ratio = 4 :1}}}}}$

Answer 2

See the Answer in Attachment

Hope It helps