Self Studies

Nuclei Test - 25

Result Self Studies

Nuclei Test - 25
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    The binding energy per nucleon of $$C^{12}$$ is $$7.68$$ MeV and that of $$C^{13}$$  is $$7.47$$ Mev. The energy required to remove one neutron from $$C^{13}$$ is
    Solution
    $$C^{13} \rightarrow C^{12}_1  +   ^1n$$

    $${BE)_C}^{12} = 12\times 7.68$$$$= 92.16\ MeV$$

    $${BE)_C}^{13} = 13\times 7.47$$$$= 97.11\ MeV$$

    $$Q = {BE)_C}^{12} - {BE)_C}^{12}$$

         $$= 97.11 - 92.16$$

         $$={4.95\ MeV}$$
  • Question 2
    1 / -0
    The atomic mass of $$_{7}N^{15}$$ is $$15.000108 amu$$ and that of $$_{8}O^{16}$$ is $$15.994915 amu$$. The minimum energy required to remove the least tightly bound proton is (mass of proton is $$1.007825 amu)$$
    Solution
    $$_8^{16}O \rightarrow   ^{15}_7N  +   ^1_1P$$

    $$\Delta m = [15.000108 + 1.007825 - 15.994915]$$
           $$= 0.013018 amu$$

    $$\Delta E = 0.013018\times 931.478$$ MeV
           $$={12.13\ MeV}$$
  • Question 3
    1 / -0
    If $$2g$$ of hydrogen is converted into $$1.986 gm$$ Helium in a thermonuclear reaction,  the energy released is
    Solution
    $$\Delta m = 2 - 1.986$$ $$= {0.014 gm}$$

    $$\Delta E = (\dfrac{0.014}{1000}Kg)\times (3\times 10^8)^2$$$$= 126\times 10^{10} J$$
  • Question 4
    1 / -0
    $$_{7}N^{14} +_{2}He^{4}\rightarrow $$ X+$$_{1}H^{1}$$ ;  X is
    Solution
    $$_7N^{14} +  _2He^4 \rightarrow   _Z^A{X} +  _1H^1$$

    Change balance: $$7 + 2 = Z + 1$$
                                         $$Z = 8$$
    Mass balance: $$14 + 4 = A + 1$$
                                      $$A = 17$$
  • Question 5
    1 / -0
    Bombardment of lithium with proton gives rise to the following reaction :
    $$^{7}_{3}Li+^{1}_{1} H \rightarrow $$2($$^{4}_{2}He) + Q $$
    The Q-value is 
    (atomic masses of lithium, proton and helium are $$7.016$$ amu, $$1.008$$ amu and $$4.004$$ amu respectively)
    Solution
    $$^7_3{Li} +   ^1_1H \rightarrow 2   ^4_2{He} + Q$$

    Mass defect $$= (7.016 + 1.008 - 2\times 4.004)\mu$$$$= 0.016 amu$$

    $$1amu = 931.478 \dfrac{MeV}{C^2}$$

    $$E = 0.016\times 931.478 \dfrac{MeV}{C^2}\times C^2$$$$=14.903 MeV$$
  • Question 6
    1 / -0
    The masses of neutron and proton are $$1.0087$$ and $$1.0073$$ amu respectively. If the neutrons and protons combine to form a Helium nucleus of mass $$4.0015 amu$$, the binding energy of the Helium nucleus will be :
    Solution
    Mass of neutron $$ = 1.0087 amu $$

    Mass of proton $$= 1.0073 amu $$

    In a helium nucleus,

       protons $$ = 2  $$

       neutrons$$ = 2     $$

    Mass(theoritically) $$ = (2\times1.0087 + 2\times1.0073)$$

                                   $$ =4.032 $$

    Mass defect $$ = (4.032- 4.0015)\ amu $$

                          $$=0.0305\  amu$$

    So, $$1\ amu$$ gives $$931.5\  MeV $$ of energy.

    So, $$0.0305\times931.5 MeV $$ of binding energy $$ = 28.4 \ MeV$$
  • Question 7
    1 / -0
    A  $$\gamma $$ -ray photon creates an electron-positron pair. The rest mass equivalent of electron is 0.5 MeV. KE of the electron - positron system is 0.78 MeV. Then the energy of $$\gamma $$ -ray photon is
    Solution
    The rest mass of electron is $$= 0.5\  MeV$$

    As, the momentum is conserved the rest mass of positron should also be $$ 0.5\  MeV$$.

    The kinetic energy of the pair is $$0.78\ MeV$$.

    $$\therefore$$ The energy of $$\gamma$$ ray photon is $$= (0.5+0.5+0.78) \ MeV= 1.78 \ MeV$$
  • Question 8
    1 / -0
    Assertion (A) : All the radioactive elements are ultimately converted into lead
    Reason (R): All the elements above lead are unstable
    Solution
    For all $$4n, 4n+2$$ series elements, on emission of different kind of rays, they form different isotopes of lead and also all element above lead are unstable .
    But, $$4n+1$$ series elements (Neptunium series), end in $$^{209}_{83}Bi $$ (Bismuth not lead)
  • Question 9
    1 / -0
    Assertion (A): Binding energy per nucleon is the measure of the stability of the nucleus.
    Reason (R): Binding energy per nucleon is more for heavier nuclides.
    Solution
    The binding energy is the energy that is released when a nucleus is assembled from its constituent nucleons  Hence, the energy equivalent of the mass-defect is called the binding-energy of the nucleus.
    The heavier the nucleus, the greater the internal repulsive forces due to the greater number of protons and less energy must be applied to remove a nucleon from the nucleus, hence the binding energy is lower. Thus for lighter nuclei binding energy is more. The greater the binding energy, the more stable the nucleus is. 
  • Question 10
    1 / -0
    In the nuclear fusion reaction : $$^{2}_{1}H+ ^{3}_{1}H \rightarrow $$ $$^{4}_{2}He+n $$ , given that the repulsive potential energy between the two nuclei is $$\sim 7.7 \times 10^{-14}$$J, the temperature to which the gases must be heated to initiate the reaction is nearly 
    (Boltzmann's constant $$k=$$1.38 x 10$$^{-23}$$J)
    Solution
    The repulsive potential energy $$=\dfrac{3}{2}kT$$

    $$7.7\times 10^{-14}=\dfrac{3}{2}\times 1.38\times\times 10^{-23} \times T$$

    $$\rightarrow \ T = 3.7\times 10^9 K \sim 10^9 K$$
Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now