Question

the ratio of difference of closest approaches of proton and an alpha particle projected towards the same nucleus with same initial kinetic energy​

Answer 1

Answer:

0.5 explanation is in the image

Answer 2

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,

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