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

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

A

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

B

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

C

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

D

`""_(18)Ar^(38)`

Text Solution

Verified by Experts

The correct Answer is:
A
Isobars have same atomic mass but different atomic number. Thus, the isobar of `""_(20)C^(40)` is `""_(18)Ar^(40)` .
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Isobars

Give one example of each of the following: (i) Isotope of ._(17)^(35)Cl (ii) Isobar of ._(18)^(40)Ar (iii) Isotone of ._(6)^(14)C

Knowledge Check

  • An isobar of ""_(20)^(40)Ca is

    A
    `""_(18)^(40)Ar`
    B
    `""_(20)^(38)Ca`
    C
    `""_(20)^(42)Ca`
    D
    `""_(19)^(39)K`

  • 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)`

  • If m_(p) is the mass of proton. m_(n) that of a neutron, M_(1) that of _(10)Ne^(20) nucleus and M_(2) that of _(20)Ca^(40) nucleus, then which of the following relations is//are not true?

    A
    `M_(2) gt 2M_(1)`
    B
    `M_(2) lt 20 (m_(p)+m_(n))`
    C
    `M_(2)=2M_(1)`
    D
    `M_(2) lt 2M_(1)`

    • Similar Questions

    Explore conceptually related problems

    Give examples of each of the following : (i) Isotope of ._(17)^(35)CI , (ii) Isobar of ._(18)^(40)Ar , (iii) Isotone of ._(7)^(15)N

    Calculate the binding energy per nucleon of ._(20)Ca^(40) nucleus. Mass of (._(20)Ca^(40)) = 39.962591 am u .

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

    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):}

    ._(8)(O)^(16) and ._(8)(O)^(18) are isotpes while ._(20)(Ca)^(40) and ._(18)(Ca)^(40) are isobars.

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