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Nuclei Test - 65

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Nuclei Test - 65
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  • Question 1
    1 / -0
    An electron collides with a fixed hydrogen atom in its ground state. Hydrogen atom gets excited and the colliding electron loses ail its kinetic energy. Consequently the hydrogen atom may emit a photon corresponding to the largest wavelength of the Balmer series. The min. K.E. of colliding electron will be
    Solution
    The electron hits the hydrogen atom in ground state with energy KK.
    This energy goes in exciting the atom. The electron loses all its energy.
    Then, the atom emits a photon by jumping back to ground state.
    So, the energy of the electron is equal to the energy of the photon emitted.
    Now the maximum wavelength in the Baimer series is for the transition between 323\to 2 and so we get 
    1λm=R(122132)=R(1419)=536R\dfrac {1}{\lambda_m} =R \left (\dfrac {1}{2^2} -\dfrac {1}{3^2}\right) =R \left (\dfrac {1}{4}-\dfrac {1}{9} \right)=\dfrac {5}{36}R
    Now, the energy of the photon emitted is
    E=hcλm=hc5R36=6.63×1034×3×108×5×1.097×10736E=\dfrac {hc}{\lambda_m}=\dfrac {hc5R}{36}=\dfrac {6.63 \times 10^{-34}\times 3\times 10^8 \times 5\times 1.097 \times 10^7}{36}
     E=3.03×1019 J\therefore \ E=3.03\times 10^{-19}\ J
     E=3.03×10191.6×1019=1.9 eV\therefore \ E=\dfrac {3.03\times 10^{-19}}{1.6\times 10^{-19}}=1.9\ eV
    Hence, the correct option is (B)(B)
  • Question 2
    1 / -0
    P and Q are two elements which form P2Q3{ P }_{ 2 }{ Q }_{ 3 } and PQ2{ PQ }_{ 2 }. If 0.15 mole of P2Q3{ P }_{ 2 }{ Q }_{ 3 } weight 15.9 g and 0.15mole of PQ2{ PQ }_{ 2 } weight 9.3 g. what are atomic weights of P and Q respectively?
    Solution
    Using- no. of mole =weightmolecular weight=\cfrac{\text {weight}}{\text{molecular weight}}

    P2Q3P_2Q_3 \Rightarrow 2P+3Q=15.9/0.15=1062P + 3Q =15.9/0.15= 106....(1)

    PQ2PQ_2 \Rightarrow  P+2Q=9.3/0.15=62P + 2Q = 9.3/0.15=62........(2)

     On solving eqn. (1) & (2) we get- 

    P=26,Q=18P =26, Q = 18
  • Question 3
    1 / -0
    92U238_{92}U^{238} on absorbing a neutron goes over to 92U239_{92}U^{239}. This nucleus emits an electron to go over to neptunium which on further emitting an electron goes over to plutonium. The plutonium nucleus can be expressed as:
    Solution
    Given 92U238 on absorbing neutron 92U239\text{Given $-*_{92} U^{238}$ on absorbing neutron $\longrightarrow _{92} U^{239}$}
    Now the nucleus -\text{Now the nucleus -}
    $$_{92}{U^{239}} \quad \underline{\text { emits an } e^{-}} \quad _{93} N p^{239}$$

    $$\begin{array}{l}\text { On further emitting an electron, } \\\qquad _{93}Np^{239} \quad \underline{\text { emits an } e^{-}} \quad _{94}Pu^{239} \\\text { Since, the emission of an } e^{-}\text{increases the atomic number } \\\text { by (1) and no change in the atomic mass. } \\\Rightarrow \text { opt (a) is correct. }\end{array}$$
  • Question 4
    1 / -0
    A nucleus ZXA_Z{X}^A emits 9α9 \alpha-particles and 5p5p particle. The ration of total protons and neutrons in the final nucleus is:-
    Solution
    Hence, option CC is correct answer.

  • Question 5
    1 / -0
    If abX_{ a }^{ b }{ X } emits a positron, two α\alpha and two β\beta^- and in last one α\alpha is also emitted and converts in  dcY_{ d }^{ c }{ Y }, correct relation:
    Solution

  • Question 6
    1 / -0
    If the binding energy per nucleon in 37Li_{3}^{7}Li and 24He_{2}^{4}He nuclei are 5.60 MeV5.60\ MeV and 7.06 MeV7.06\ MeV respectively, then in the reaction : p+37Li224Hep + _{3}^{7}Li \rightarrow 2_{2}^{4}He energy of proton must be
    Solution
     Given BE of 37Li=5.6MeV,BE of 24He=7.06MeVp+73Li242He\begin{array}{l}\text { Given }-B E \text { of }_{3}^{7} L_{i}=5.6\mathrm{MeV} , B E \text { of }_{2}^{4}\mathrm{He}=7.06\mathrm{MeV} \\p+\frac{7}{3}\mathrm{Li}\rightarrow 2 \cdot \frac{4}{2} \mathrm{He}\end{array}


     Energy of proton =2BE of HeBE of Li =2[4×7.06][7×5.60]=17.28MeV\begin{aligned}\text { Energy of proton } &=2 \cdot B E \text { of }\mathrm{He}-\mathrm{BE} \text { of Li } \\&=2 \cdot[4 \times 7.06]-[7\times 5.60] \\&=17.28 \mathrm{MeV}\end{aligned}
  • Question 7
    1 / -0

    For a nucleus to be stable, the correct relation between neutron number N and proton number Z is 

    Solution

  • Question 8
    1 / -0
    Let uu denote one atomic mass unit. One atom of an element of mass number AA has mass exactly equal to AuAu
    Solution

  • Question 9
    1 / -0
    If half-life of a radioactive substance is 6060 minutes, then the percentage decay in 44 hours is?
    Solution

  • Question 10
    1 / -0
    In a hydrogen atom, the binding energy of the electron in the ground state is E1.E_{1}. Then the frequency of revolution of nth electron in the nth orbits is
    Solution
    B.E of electron= K.E of electron (in that orbit)12mv2=E1(i)angular momentum -mvr=nh2π(ii)Now from (i) divided by (ii)v2r=E1×2πnhv2πr=2E1nh    f=2E1nh\begin{array}{l}\text{B.E of electron= K.E of electron (in that orbit)}\\\dfrac{1}{2}mv^2=E_1-(i)\\\therefore \text{angular momentum -} mvr = \dfrac{nh}{2 \pi}-(ii)\\\text{Now from (i) divided by (ii)}\\\dfrac{v}{2r}= \dfrac{E_1 \times 2 \pi}{nh}\\\dfrac{v}{2 \pi r}=\dfrac{2E_1}{nh}\\\implies f=\dfrac{2E_1}{nh}\end{array}
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