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A neutron of energy 1 MeV and mass 1.6 x...

A neutron of energy 1 MeV and mass `1.6 xx 10^(-27)` kg passes a proton at such a distance that the angular momentum of the neutron relative to the proton approximately equals `10^(-33) J s`. The distance of closest approach neglecting the interaction between particles si

A

`0.44` nm

B

0.44 mm

C

0.44 Å

D

0.44 fm

Text Solution

Verified by Experts

The correct Answer is:
D
If d is the distance of closedt approach given , then the angular mometum = mvd = `10^(-33) Js`
`E = (1)/(2) mv^(2) = 1 Me V = 1.6 xx 10^-13` J
Momentum ,
`P = sqrt(2 m_(n) E) = sqrt(2 xx 1.6 xx 10^(-27) xx 1.6 xx 10^(-13)) = 1.6 sqrt2 xx 10^(-20) kg m s^(-1)`
Distance of closest approach ,
`d = (10^(-33))/(1.6 sqrt2 xx 10^(-20))`
`(1)/(1.6 sqrt2) xx 10^(-13) = (100)/(1.6 sqrt2) fm = 0.44` fm
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Knowledge Check

  • A proton has a mass of 1.673xx10^(-27)kg . If the proton is accelerated to a speed of 0.93c , what is the magnitude of the relativistic momentum of the proton?

    A
    `6.2xx10^(-17)kg*m//s`
    B
    `4.7xx10^(-19)kg*m//s`
    C
    `1.3xx10^(-18)kg*m//s`
    D
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  • A neutron having a mass of 1.67 xx 10^(-27) kg and moving at 10^(8) m//s collides with a deuteron at rest and sticks to it. If the mass of the deuteron is 3.33 xx 10^(-27) kg then the speed of the combination is

    A
    `2.56 xx 10^(3) m//s`
    B
    `2.98 xx 10^(5) m//s`
    C
    `3.33 xx 10^(7) m//s`
    D
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  • Rest mass of 1 mole neutrons (m_(n)=1.675xx10^(-27)kg) is:

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    B
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    C
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    D
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