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An alpha-particle of kinetic energy 7.68...

An `alpha`-particle of kinetic energy 7.68 MeV is projected towards the nucleus of copper (Z=29). Calculate its distance of nearest approach.

A

`8.4xx10^(-15)cm`

B

`8.4xx10^(-15)m`

C

`4.2xx10^(-15)m`

D

`4.2xx10^(-15)cm`

Text Solution

Verified by Experts

The correct Answer is:
B
Let the minimum distance of approach be `r_(0)`. At this distance, the whole of the kinetic energy of the alpha - particle will be converted into the electrical potential energy.
The position charge on the `alpha-` particle = 2e.
`10xx(1.6xx10^(-13))=(1)/(4piepsilon_(0))((29e)(2e))/(r_(0))`
`therefore r_(0)=((9xx10^(9))(29)/(2)(1.6xx10^(-19))^(2))/(10xx(1.6xx10^(-13)))=8.4xx10^(-15)m`
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MTG GUIDE-ATOMS AND NUCLEI-NEET CAFÉ (TOPICWISE PRACTICE QUESTIONS)(ALPHA - PARTICLE SCATTERING EXPERIMENT AND RUTHERFORD.S MODEL OF ATOM)
  1. An alpha-particle of kinetic energy 7.68 MeV is projected towards the ...

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  2. A beam of beryllium nucleus (z = 4) of kinetic energy 5.3 MeV is heade...

    Text Solution

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