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Dual Nature of Radiation and Matter Test - 23

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Dual Nature of Radiation and Matter Test - 23
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  • Question 1
    1 / -0
    Choose the correct statement
    Solution
    Propagation of de Broglie waves in 1d real part of the complex amplitude is blue, imaginary part is green. The probability (shown as the colour opacity) of finding the particle at a given point x is spread out like a waveform, there is no definite position of the particle. As the amplitude increases above zero the curvature decreases, so the amplitude decreases again, and vice versa the result is an alternating amplitude: a wave. So, the matter waves are waves of probability amplitude

  • Question 2
    1 / -0
    Which of the following particles - neutron, proton,electron and deuteron has the lowest energy if all have the same de Broglie wavelength:
    Solution
    $$\lambda =\dfrac{h}{mv}$$

    $$\lambda =\dfrac{h}{\sqrt{2Em}}$$

    So  $$E\alpha \dfrac{1}{m}$$
    Deuteron is mass of 1 proton and 1 neutron ,hence its mass is maximum out of the given particles ,so its energy is the lowest.
  • Question 3
    1 / -0
    In photo-electric effect, when photons of energy $$'hv\ '$$ are incident on a metal surface, electrons are emitted with some kinetic energy. It is possible to say that
    Solution
    $$hv=hv_{0}+K.E.$$   for emitted electrons, its possible that they have a K.E from zero to $$hv-hv_{0}$$. Hence correct option is B.
  • Question 4
    1 / -0
    An electron of charge 'e' and mass 'm' is accelerated from rest by a potential difference 'V'. The de Broglie wavelength is
    Solution
    $$\frac{1}{2}mv^{2}=eV$$
    $$mv=\sqrt{2eVm}$$
    So de-broglie wavelength
    $$\lambda =\dfrac{h}{mv}$$
       $$=\dfrac{h}{\sqrt{2meV}}$$
    $$\therefore \lambda \alpha \frac{1}{\sqrt{V}}$$
  • Question 5
    1 / -0
    Two photons, each of energy 2.5 eV are simultaneously incident on the metal surface. If the work function of the metal is 4.5 eV, then from the surface of the metal
    Solution
    By definition  the work function is the minimum energy (in electron volts) needed to remove an electron from a solid to a point immediately outside the solid surface (or energy needed to move an electron from the Fermi level into vacuum).
    Hence, the work function of metal is 4.5 eV this means that the energy necessary to emit electrons from metal surface is 4.5 eV. But the incident photons are of energy 2.5 eV, which is not sufficient to emit electrons from metal surface hence no electron will emit from the metal surface.
    So, the answer is option (B)..
  • Question 6
    1 / -0
    The graph between the de Broglie wavelength and the momentum of a photon is a 
    Solution
    The relationship between de Broglie wavelength $$(\lambda)$$ and the momentum (p) of a photon is given by:
    $$\lambda=\dfrac{h}{p}$$.
    So, the graph between the de Broglie wavelength and the momentum of a photon is a rectangular hyperbola.
  • Question 7
    1 / -0
    A cathode ray particle is accelerated from rest through a potential difference of 'V' volt. Speed of the particle is
    Solution
    W = eV = KE
    mass of the particle = m
    speed of the particle = $$v$$

    $$\frac{1}{2} mv^2 = eV$$
    So, $$v$$ = $$ \sqrt{\dfrac{2eV}{m}}$$
  • Question 8
    1 / -0
    The ratio of the wavelengths of a photon and that of an electron of same energy $$E$$ will be [$$m$$ is mass of electron]:
    Solution
    Wavelength of photon:
    $$E=\dfrac{hc}{\lambda }$$

    $$\lambda _{p}=\dfrac{hc}{E}$$

    Wavelength of electron:
    $$\lambda _{e}=\dfrac{h}{mv}$$

    $$\dfrac{1}{2}mv^2=E$$

    $$mv=\sqrt{2Em}$$

    So  $$\lambda _{e}=\dfrac{h}{\sqrt{2Em}}$$

    $$\dfrac{\lambda _{p}}{\lambda _{e}}=c\sqrt{\dfrac{2m}{E}}$$
  • Question 9
    1 / -0
    The wavelengths of a proton and a photon are same. Then :
    Solution

    For proton,

    $$\lambda =\dfrac{h}{mv}$$

    $$\lambda _{1}=\dfrac{h}{p}$$

    For photon,

    $$E=\dfrac{hc}{\lambda }$$

    $$\lambda _{2}=\dfrac{h}{E/c}$$

    &  $$\lambda _{1}=\lambda _{2}$$

    So  $$p=\dfrac{E}{c}$$
  • Question 10
    1 / -0
    An electron and a proton possess the same amount of Kinetic energy. Then the relation between the wavelength of electron$$(\lambda _{e})$$ and the wavelength of proton$$(\lambda _{p})$$ is
    Solution
    For an electron
    $$\dfrac{1}{2}mv^{2}=E$$

    $$mv=\sqrt{2Em}$$

    $$\lambda _{e}=\dfrac{h}{mv}$$

    $$=\dfrac{h}{\sqrt{2Em_{e}}}$$

    For a proton

    $$\dfrac{1}{2}mv^{2}=E$$

    $$mv=\sqrt{2Em}$$

    $$\lambda _{p}=\dfrac{h}{\sqrt{2Em_{p}}}$$

    $$\dfrac{\lambda _{e}}{\lambda _{p}}=\sqrt{\dfrac{m_{p}}{m_{e}}}$$
    $$\dfrac{\lambda _{e}}{\lambda _{p}}> 1$$
    $$\lambda _{e}> \lambda _{p}$$
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