Self Studies

Nuclei Test - 46

Result Self Studies

Nuclei Test - 46
  • 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
    Which symbol is used to represent the unit of atomic mass, amu?
    Solution
    The symbol 'u' represents the unit of atomic mass, amu. For example, the atomic mass of carbon-12 is written as 12 u. 
    Hence, option A is correct.
  • Question 2
    1 / -0
    A nuclear transformation is denoted by $$X(n, \alpha) \rightarrow _{3}^{7}Li$$. Which of the following is the nucleus of element $$X$$?
    Solution
    $$\begin{aligned}& X(n, \alpha) \longrightarrow{ }_{3}^{7} L_{i} \\X &+n \quad \longrightarrow \quad{ }_{3}^7 L_{i}+\alpha \\\alpha &={ }_2^{4} H_{e}\\n&=^0n\end{aligned}$$

    $$\begin{array}{l}X+{ }^{1} n \longrightarrow  _{3}^{7} Li+{ }^{4}_2 H_{e} \\X \text { must have Atomic mass }=10\end{array}$$
    $$\text { and Atomic no. }=5$$

    $${ }^{10}_5 X={ }^{10}_5 B$$
  • Question 3
    1 / -0
    The binding energies per nucleon of deuteron $$(_{1}H^{2})$$ and helium atom $$(_{2}He^{4})$$ are $$1.1\ MeV$$ and $$7\ MeV$$. If two deuteron atoms react to form a single helium atom, then the energy released is
    Solution
    ATQ, following reaction takes place,
    $${ H }_{ 1 }^{ 2 }+{ H }_{ 1 }^{ 2 }\rightarrow { H }_{ 2 }^{ 4 }\quad -Q$$
    {-Q: energy reloaded}
    $$\rightarrow 2({ H }_{ 1 }^{ 2 })\rightarrow { H }_{ 2 }^{ 4 }\quad -Q\\ $$
    So, ATQ binding energy per nucleon of deuteron=1.1MeV.
    Therefore, There are 2 nucleon in deuteron atom, hence total binding energy
    $$=1.1\times 2\\ =2.2MeV$$
    Therefore, for left hand part, BE of
    $${ H }_{ 1 }^{ 2 }=2.2MeV$$
    Similarly, BE per nucleon of helium atom= 7eV.
    Number of nucleons in helium$$=4$$
    Total BE of nucleus$$==7\times 4=28MeV$$
    Now, substituting the values of BE in the equation to get the value of Q,
    $$\Rightarrow 2(2.2)\rightarrow 28-Q\\ \rightarrow Q=28-4.4=23.6MeV$$


  • Question 4
    1 / -0
    A radioactive nuclide X decays into nuclei T and Z by simultaneous disintegration as shown. Effective decay constant for the disintegration is

    Solution

    $$\begin{array}{l}\frac{d y}{d t}=\lambda_{1} x---\text { for } x \rightarrow y \\\frac{d z}{d t}=\lambda_{2} x--- \text { for } x \rightarrow z \\\text { Now adding }\\\left(\lambda_{1}+\lambda_{2}\right) x=\frac{d y}{d t}+\frac{d z}{d t}\longrightarrow(i) \\\frac{d y}{d t}+\frac{d z}{d t}=-\frac{d x}{d t}\longrightarrow\text { (iii) }\end{array}$$

    $$\begin{array}{l}\text { Rate of decrease of } x \\\text { Would be equal to } \\\text { net increase rale } \\\text { of } y \text { and } z \text { . } \\-\frac{d x}{d t}=\left(\lambda_1+\lambda_{2}\right)x\\\Rightarrow \lambda_{\text {effective }}=\lambda_{1}+\lambda_{2}\end{array}$$

  • Question 5
    1 / -0
    Decreasing order of atomic weight is correct of the elements given below?
    Solution
    Atomic weight of Iron $$\approx 56$$ u
    Atomic weight of Cobalt $$\approx 59$$ u
    Atomic weight of Nickel $$\approx 58.7$$ u
    So, the order of atomic weights is $$Co>Ni>Fe$$.
    Option $$C$$ is correct.
  • Question 6
    1 / -0
    A collective name for nucleons and other elementary particles that decay into nucleons by the emission of mesons is?
    Solution
    $$\begin{array}{l}\text { A collective name for } \\\text { nucleons and other elementry } \\\text { particles that decay into } \\\text { nucleons by the emission } \\\text { of mesons is Hadron. }\end{array}$$
  • Question 7
    1 / -0
    The mass of $$_{ 17 }{ { CI }^{ 35 } }$$ is $$34.9800$$ amu. Calculate its binding energy (mass of $$_{ 0 }{ { n }^{ 1 } }=1.008665$$ amu $$_{ 1 }{ { H }^{ 1 } }=1.007825$$ amu):
    Solution
    There are 17 protons  and 18 neutrons in the nucleus of chlorine
    $$\Delta$$ m=Mass of the nucleons combined- Actual mass of the nucleus  
    $$=17\times 1.0007825+18\times 1.0086-34.9800$$
    $$=35.288995-34.98$$
    $$=0.308995$$ 
    B.E.$$=\Delta m(931.5)=279.45$$
  • Question 8
    1 / -0
    The relation between the volume $$V$$ and the mass $$M$$ of a nucleus is:
    Solution
    Since the density of nucleus is fixed.
    $$D=\cfrac { M }{ V } \\ \Rightarrow M=DV\\ M\propto V$$
  • Question 9
    1 / -0
    The above is a plot of binding energy per nucleon $${E}_{b}$$, against the nuclear mass $$M$$; $$A,B,C,D,E,F$$ correspond to different nuclei. Consider four reactions:
    (i) $$A+B\rightarrow C+\varepsilon $$
    (ii) $$C\rightarrow A+B+\varepsilon $$
    (iii) $$D+E\rightarrow F+\varepsilon $$
    (iv) $$F\rightarrow D+E+\varepsilon $$
    where $$\varepsilon $$ is the energy released? In which reactions is $$\varepsilon $$ positive. 

    Solution
    In a spontaneous reaction product has always Binding energy greater than that of reactants. So, (i) is a fusion reaction and (iv) is a fission reaction.
  • Question 10
    1 / -0
    A nucleus of mass $$M+\Delta m$$ is at rest and decays into two daughter nuclei of equal mass $$\cfrac{M}{2}$$ each. Speed of light is $$c$$. The binding energy per nucleon for the parent nucleus is $${E}_{1}$$ and that for the daughter nuclei is $${E}_{2}$$. Then :
    Solution
    $$ \begin{array}{l} \text { let } x \rightarrow 2 y \\ \text { mass } \rightarrow M+\Delta m \quad 2 \times \frac{M}{2}=M \\ \Delta m \text { mass is lost as energy } \\ \Rightarrow \text { energy is released } \\ \Rightarrow \text { energy of } x>\text { energy of y } \\ \qquad E_{1}>E_{2} \\ \text { option } B \end{array} $$
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