Nuclei

Nuclei

Easy Questions

Modern Physics

Dual Nature of Radiation and Matter
Atoms
Nuclei
Semiconductor Electronics : Materials, Devices and Simple Circuits
  1. The nuclei 6C13 and 7N14 can be described as








  1. View Hint View Answer Discuss in Forum

    As 6C13 and 7N14 have same no. of neutrons (13 – 6 =7 for C and 14 – 7 =7 for N),
    so they are isotones.

    Correct Option: A

    As 6C13 and 7N14 have same no. of neutrons (13 – 6 =7 for C and 14 – 7 =7 for N),
    so they are isotones.

  1. The ratio of the radii of the nuclei 13Al27 and 52Te125 is approximately​​








  1. View Hint View Answer Discuss in Forum

    R ∝ (A)1/3
    ∴ RAl ∝ (27)1/3 and RTe ∝ (125)1/3

    ∴
    RAl
    		=
    3
    		=
    6
    RTe510

    Correct Option: A

    R ∝ (A)1/3
    ∴ RAl ∝ (27)1/3 and RTe ∝ (125)1/3

    ∴
    RAl
    		=
    3
    		=
    6
    RTe510

  1. A nucleus of uranium decays at rest into nuclei of thorium and helium.
    Then :​​








  1. View Hint View Answer Discuss in Forum

    In an explosion a body breaks up into two pieces of unequal masses both part will have numerically equal momentum and lighter part will have more velocity.
    U → Th + He

    KETh
    P2
    		, KEHe =
    P2
    2mTh2mHe

    since mHe is less so KETh will be more.

    Correct Option: D

    In an explosion a body breaks up into two pieces of unequal masses both part will have numerically equal momentum and lighter part will have more velocity.
    U → Th + He

    KETh
    P2
    		, KEHe =
    P2
    2mTh2mHe

    since mHe is less so KETh will be more.

  1. The Binding energy per nucleon of 3Li7 and 2He4 nuclei are 5.60 MeV and 7.06 MeV, respectively. ​
    In the nuclear reaction 3Li7 + 1H1 → 2He4 + Q, the value of energy Q released is :​








  1. View Hint View Answer Discuss in Forum

    BE of  2He4 = 4 × 7.06 = 28.24 MeV

    BE of
    7
    		Li = 7 × 5.60 = 39.20 MeV
    3


    Therefore, Q = 56.48 – 39.20 = 17.28 MeV.

    Correct Option: D

    BE of  2He4 = 4 × 7.06 = 28.24 MeV

    BE of
    7
    		Li = 7 × 5.60 = 39.20 MeV
    3


    Therefore, Q = 56.48 – 39.20 = 17.28 MeV.

  1. A sample of radioactive element has a mass of 10 gm at an instant t=0. The approximate mass of this element in the sample after two mean lives is








  1. View Hint View Answer Discuss in Forum

    Using the relation for mean life.

    Given : t = 2τ = 2
    1
    λ

    ∴ τ =
    1
    λ

    Then from
    = 10
    1
    		2= 1.35g
    2

    Correct Option: B

    Using the relation for mean life.

    Given : t = 2τ = 2
    1
    λ

    ∴ τ =
    1
    λ

    Then from
    = 10
    1
    		2= 1.35g
    2