In a head-on collision between an alpha-particle and a gold nucleus, the distance of closest approach is 39.5 fermi. Calculate the energy of alpha-particle. For gold Z=79.

In a head-on collision between an alpha-particle and a gold nucleus, t

1.	In a head-on collision between an alpha-particle and a gold nucleus, the distance of closest approach is 39.5 fermi. Calculate the energy of alpha-particle. For gold Z=79.
Answer» <html><body><p> <br/> <br/></p>Solution :Given `r_(0)=39.5 "fm"=39.5xx10^(-15)m, =79, <a href="https://interviewquestions.tuteehub.com/tag/e-444102" style="font-weight:bold;" target="_blank" title="Click to know more about E">E</a>=?` <br/> As `E=(Ze(2e))/(4 pi epsilon_(0)<a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>)=(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> Ze^(2))/(4 pi epsilon_(0)r)` <br/> `=(9xx10^(9)xx79xx2xx(1.6xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/19-280618" style="font-weight:bold;" target="_blank" title="Click to know more about 19">19</a>))^(2))/(39.5xx10^(-15))` <br/> `=(9xx79xx2xx1.6xx1.6xx10^(-14))/(39.5)J` <br/> or `E=(18xx79xx1.6xx1.6xx10^(-14))/(39.5xx1.6xx10^(-13))` <br/> `=5.76` M eV.</body></html>

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In a head-on collision between an alpha-particle and a gold nucleus, the distance of closest approach is 39.5 fermi. Calculate the energy of alpha-particle. For gold Z=79.

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