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The nuclear mass of .26F^(56) is 55.85u....

The nuclear mass of `._26F^(56)` is 55.85u. Calculate its nuclear density.

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To calculate the nuclear density of fluorine-56 (represented as \( _{26}^{56}F \)), we will follow these steps: ### Step 1: Convert Nuclear Mass to Kilograms The nuclear mass of fluorine-56 is given as 55.85 u. We need to convert this mass into kilograms. The conversion factor from atomic mass units (u) to kilograms is approximately \( 1 \, \text{u} = 1.66 \times 10^{-27} \, \text{kg} \). \[ \text{Mass in kg} = 55.85 \, \text{u} \times 1.66 \times 10^{-27} \, \text{kg/u} \] ...
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Knowledge Check

  • Assume that the nuclear mass is of the order of 10^(-27) kg and the nuclear radius is of the order of 10^(15)m . The nuclear density is of the order of

    A
    `10^(2)Kg//m^(3)`
    B
    `10^(10)Kg//m^(3)`
    C
    `10^(17)Kg//m^(3)`
    D
    `10^(31)Kg//m^(3)`
  • Given the mass of iron nucleus as 55.85u and A = 56, the nuclear density is (u = 1.66 xx 10 ^(-27) kg, r = 1.2 xx 10 ^(-15)m )

    A
    `1.29 xx 10 ^(-7)kgm^(-3)`
    B
    `2.29 xx 10 ^(17 kgm ^(-3))`
    C
    `2.29 xx 10 ^(-7 kgm ^(-3))`
    D
    `1.29 xx 10 ^(-27 kgm ^(-3))`
  • The nuclear radius of ._(8)O^(16) is 3 xx10^(-15) m . If an atomic mass unit is 1.67 xx 10^(-27) kg , then the nuclear density is approximately.

    A
    `2.35 xx 10^(17) g cm^(-3)`
    B
    `2.35 xx 10^(17) kg m^(-3)`
    C
    `2.35 xx 10^(17) g m^(-3)`
    D
    `2.35 xx 10^(17) kg mm^(-3)`
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