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Assertion (A) : The positively charged n...

Assertion (A) : The positively charged nucleus of an atom has a radius of almost `10^(-15)` m.
Reason (R) : In a-particle scattering experiment the distance of closest approach for `alpha` -particles is of the order of `10^(-15)` m.

A

If both assertion and reason are true and the reason is the correct explanation of the assertion.

B

If both assertion and reason are true but reason is not the correct explanation of the assertion.

C

If assertion is true but reason is false.

D

If the assertion is false but reason is true.

Text Solution

Verified by Experts

The correct Answer is:
A
Distance of closest approach provides an upper limit for the size of nucleus.
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Knowledge Check

  • Assertion: The positively changed nucleus of an atom has a radius of almost 10^(-15)m . Reason: In alpha -particle scattering experiment the distance of closest apporach for alpha -particles is ~~10^(-15)m .

    A
    If both assertion and reason are true and reason is the correct explanation of assertion.
    B
    If both assertion and reason are true but reason is not the correct explanation of assertion.
    C
    If assertion is true but reason is false.
    D
    If assertion is false but reason is true.

  • In scattering experiment, find the distance of closest approach, if a 6 MeV alpha - particle is used

    A
    `3.2 xx 10^(-16) m`
    B
    `2 xx 10^(-14) m`
    C
    `4.6 xx 10^(-15) m`
    D
    `3.2 xx 10^(-15)` m

  • In mu-particle scattering experiment, the expression of distance of closest approach ( r_0 )is (symbols have their usual meanings)

    A
    `1/(4πepsilon_0) × (Ze^2)/E`
    B
    `1/(4πepsilon_0) × (2Ze^2)/E`
    C
    `1/(4πepsilon_0) × (Z^2e^2)`
    D
    `1/(4πepsilon_0)×(Z^2e^4)/E`

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    A n alpha -particle of velocity 1.6xx10^(7) m s^(-1) approaches a gold nucleii (Z=79) . Calculate the distance of closest approach. Mas of an alpha -particle =6.6xx10^(-27) kg .

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