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Atoms Test - 25

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Atoms Test - 25
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Weekly Quiz Competition
  • Question 1
    1 / -0
    When the particle and its anti-particle unite, the result is
  • Question 2
    1 / -0
    In the Geiger-Marsden experiment, the force that scatters particles is 
    Solution
    Coulomb force is responsible for scattering of particles. When the alpha particles (positive in charge) get closer to the nucleus, which is positive in charge, they get repelled through various angles.
  • Question 3
    1 / -0
    The force experienced by the cathode rays when they pass through a uniform electric field of intensity $$\bar{E}$$ is:
    Solution
    Cathode particles are negatively charged
    $$\therefore F_E = -q \vec{E} $$
    Force is opposite to the direction of electric field
  • Question 4
    1 / -0
    Cathode rays are made to pass between the poles of a magnet as shown in figure. The effect of magnetic field is

    Solution
    Force will be perpendicular and outside to the plane of the paper

  • Question 5
    1 / -0
    The angular momentum of the $$\alpha -$$particles which are scattered through large angles by the heavier nuclei, is conserved because
    Solution

    The angular momentum of $$\alpha$$ particles is conserved because there is no external torque.

  • Question 6
    1 / -0
    An electron makes transition from $$n= 3 , n=1$$ state in a hydrogen atom. The different possible number of photons that can be emitted is :
    Solution
    For transition from $$n=3$$ to $$n=1$$
    Possible number of photons $$=\ ^3C_2$$$$=\dfrac{3!}{(3-2)!2!}$$$$=\dfrac{3\times 2\times 1}{1\times 2\times 1}=3$$
  • Question 7
    1 / -0
    In the lowest energy level of hydrogen atom, electron has an angular momentum equal to:
    Solution
    The angular momentum$$ L$$ $$=m_{e}vr$$ is on integer multiple of $$\dfrac{h}{2\pi }$$

    $$mvr=\dfrac{nh}{2\pi}$$ 

    For, $$n=1$$

    $$mvr=\dfrac{h}{2\pi}$$
  • Question 8
    1 / -0
    The ionization energy of a hydrogen-like ion $$A$$ is greater than that of another hydrogen-like ion $$B$$. Let $$u$$ and $$E$$ represent the speed of the electron and energy of the atom respectively in the ground state. Then, 
    Solution
    We know

    $$E=-13.6\dfrac{Z^{2}}{n^{2}}eV$$

    So Ionization energy $$=13.6\dfrac{Z^{2}}{n^{2}}eV$$

    given,  $$IE_{A}> IE_{B},$$ so $$E_{A}< E_{B}$$

    that means $$Z_{A}> Z_{B}$$

    Now   $$u   \propto \dfrac{Z}{n}$$ 

            $$u_{A}> u_{B}$$
  • Question 9
    1 / -0
    In Rutherford's experiment the number of $$\alpha$$ particles scattered through an angle $$60^o$$ is $$112$$ per minute, then the number of $$\alpha$$ particles scattered through an angle of $$90^{0}$$ per minute by the same nucleus is:
    Solution
    In Rutherford's experiment the number of  particles scattered through an angle $$\theta$$ per minute is given by:
    $$N \propto \dfrac{Z^2}{Sin^4(\theta /2)}$$

    According to the question: the number of  particles scattered through an angle $$60^o$$ is 112 per minute. So,
    $$112 \propto \dfrac{Z^2}{Sin^4(60 /2)}$$------ (1)

    And, lets assume the number of  particles scattered through an angle $$90^o$$ is N' per minute
    $$N' \propto \dfrac{Z^2}{Sin^4(90 /2)}$$------(2)

    Using (1) and (2),
    $$\dfrac{N'}{112} \propto \dfrac{Sin^4(60 /2)}{Sin^4(90 /2)}$$

    $$N'=28$$
  • Question 10
    1 / -0
    The radius of hydrogen atom, when it is in its second excited state ,becomes $$\underline{\hspace{0.5in}}$$ its ground state radius. 
    Solution
    $$r=\dfrac{a_{0}n^{2}}{Z}$$
    $$r_{1}$$ ( ground state ) $$={a_{0}}(1)^{2}$$
                                    $$=a_{0}$$
    $$r_{2}$$ ( 2nd excited state ) $$=a_{0}(3)^{2}$$
                                           $$=9a_{0}$$
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