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Atoms Test - 61

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Atoms Test - 61
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  • Question 1
    1 / -0
    In a muonic atom, a muon of mass of 200 times of that of electron and same charge is bond to the proton. The wavelengths of its Balmer series are in the range of  :
    Solution
    Energy of an orbit,

    E=me4802h2z2n2E=\dfrac{-me^4}{8\in_0^2h^2}\dfrac{z^2}{n^2}

    E\because E increase 200 times

    λ\therefore \lambda will decrease by 200 times.

    Hence λ\lambda will be in the range of x-rays.
  • Question 2
    1 / -0
    A student measures the level of radiation emitted by a radioactive sample.
    The table shows the readings she records on the counter over a short period of time.
    Counter reading/ counts per minute10610696969898100100
    The sample is removed and the counter then shows a background radiation reading of 44 counts per minute.
    What is the best estimate for the average count rate due to the radioactive sample?

    Solution
     Total count per minute =106+96+98+1100=400 106 +96 + 98 +1 100 = 400 counts 
    so, Average count of sample 4004=100\dfrac {400}{4} = 100
    net count per minute = Average count - number of sample
                                         = 1004=96100 - 4 = 96 counts per minute
  • Question 3
    1 / -0
    Which statement about the structure of an atom is correct?
    Solution
    In an atom  center is called nucleus consist of proton and neutron which are positively charged and neutral respectively
     so nucleus is net positive
    and nucleus is surrounded by negatively charged electron which account s for neutrality of atom
  • Question 4
    1 / -0
    Cathode rays are made to pass between the plates of a charged capacitor, It attracts.
    Solution

  • Question 5
    1 / -0
    The cathode ray particles originates in a discharge tube from the
    Solution
    Cathode rays are so named because they are emitted by the negative electrode, or cathode, in a vacuum tube. To release electrons into the tube, they first must be detached from the atoms of the cathode

    Hence option A is correct
  • Question 6
    1 / -0
    The diagram shows a cathode-ray oscilloscope display of an electromagnetic wave. The time base setting is 0.20μscm1 0.20 \mu s cm^{-1}. Which statement is correct? 

    Solution
    It is given that
    λ2=1.0 cm\dfrac{\lambda}{2}= 1.0\ cm
        λ=2.0 cm\implies \lambda = 2.0\ cm
    time base = 0.20μ scm10.20\mu\ scm^{-1}
    So, 
    Time period = 0.20μ cms1×λ0.20\mu\ cms^{-1}\times \lambda
                     T=0.4μ s\implies T=0.4\mu\ s
    So, frequency = 1T\dfrac{1}{T}
                  f=10.40×106 s1\implies f = \dfrac{1}{0.40\times 10^{-6}}\ s^{-1}
                   f=2.5MHz\implies f = 2.5MHz
    Since the frequency range of radio range is 
    So, the given wave is in radio-wave region of spectrum.
  • Question 7
    1 / -0
    For which of the following set of quantum numbers, an electron will have the lowest energy ?
    Solution

  • Question 8
    1 / -0
    Which is the common node for all orbitals:
    Solution

  • Question 9
    1 / -0
    Which of the following curves may represent radius of orbit in HH-atom as a function quantum number ?
    Solution

  • Question 10
    1 / -0
    The number of orbitals in 3rd3^{rd} orbit are :
    Solution

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