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An isobar of ""(20)^(40)Ca is...

An isobar of `""_(20)^(40)Ca` is

A

`""_(18)^(40)Ar`

B

`""_(20)^(38)Ca`

C

`""_(20)^(42)Ca`

D

`""_(19)^(39)K`

Text Solution

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The correct Answer is:
A
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Isobars

Find the atomic structure of ._(20)^(40)Ca .

Knowledge Check

  • An isobar of ._(20)Ca^(40) is

    A
    `._(18)Ar^(40)`
    B
    `._(20)Ca^(38)`
    C
    `._(20)Ca^(42)`
    D
    `._(18)Ar^(38)`

  • The calculated mass of ""_(20)Ca^(40) is 40.328 u. It releases 306.3 MeV energy in a nuclear process . Its isotopic mass is

    A
    39.998
    B
    40.657
    C
    0.329
    D
    `2.85xx10^(4)`

  • Let m_(p) , be the mass of a proton, m_(n) , that of a neutron, M_(1) , that of a ""_(10)^(20)Ne nucleus and M_(2) , that of a ""_(20)^(40) Ca nucleus. Then

    A
    `M_(2) = 2M_(1)`
    B
    `M_(1) lt 10(m_(p) + m_(n))`
    C
    `M_(2) gt 2M_(1)`
    D
    `M_(1) = M_(2)`

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    What is the minimum photon energy required to remove the least bound neutron of ._(20)^(40)Ca and ._(18)^(40)Ar . The nesessary atomic masses (in mu ) are given below: {:(M(.^(40)Ca)=39.962591 u),(M(.^(39)Ca)=38.970719 u),(M(.^(40)Ar)=39.962383 ),(M(.^(39)Ar)=38.964314 u),(" "m_n=1.008665 u):}

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    Which of the following pairs represent isotopes and which represent isobars? ""_(19)^(40)K, ""_(20 )^(40)Ca Calculate the difference in the number of neutrons in the isotopic pairs.

    In the nucleus of ""_(20)^(40) Ca, there are

    Let m_(p) be the mass of a poton , M_(1) the mass of a _(10)^(20) Ne nucleus and M_(2) the mass of a _(20)^(40) Ca nucleus . Then

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