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Dual Nature of Radiation and Matter Test - 41

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Dual Nature of Radiation and Matter Test - 41
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  • Question 1
    1 / -0
    de-Broglie wavelength of atom at $$TK$$ absolute temperature will be
    Solution
    de-Broglie wavelength  $$\lambda = \dfrac{h}{mv}$$
    Kinetic energy of the atom  $$K.E = \dfrac{1}{2}mv^2 = \dfrac{3}{2}KT$$
    $$\implies \ mv = \sqrt{3mKT}$$
    Thus we get  $$\lambda = \dfrac{h}{\sqrt{3mKT}}$$
    So, option B is correct.
  • Question 2
    1 / -0
    According to Einsteins photoelectric equation, the graph between the kinetic energy of photoelectrons ejected and the frequency of incident radiation is :
    Solution
    The maximum kinetic energy of photoelectron ejected is given by,

    $$K.E. = h\upsilon \, - \, \phi _{0} \, = \, h\upsilon \,-\, h\upsilon _{0}$$

    Where work function depends on the type of material.

    If the frequency of incident radiation is greater than $$\upsilon _{0}$$ only then the ejection of photoelectrons start. After that as frequency increases kinetic energy also increases. 
    Option D
  • Question 3
    1 / -0
    A material particle with a rest mass $$m_0$$ is moving with a velocity of light c. Then determine the wavelength of the de Broglie wave associated with it is.
    Solution
    A
    The de Broglie wavelength is 
    $$\lambda =\frac{h}{p} = \frac{h}{mv}$$
    Where $$h$$ = planks constant 
         $$m$$ = rest mass
          $$v$$ = Velocity 
    Here $$m_{o}$$is the rest mass and $$c$$ is the velocity. So the de Broglie wavelength is 
    $$\lambda =\frac{h}{m_{o}c}$$
  • Question 4
    1 / -0
    Which of the following happens when a body is charged positively?
    Solution
    Protons are present inside the nucleus and cannot move out of it but electrons are mobile and can move from atom to atom or material to material.
    Thus positive charge occurs due to the deficiency of electrons in a material.
  • Question 5
    1 / -0
    The time taken by a photoelectron to come Out after the photon strikes is approximately.
    Solution
    Emission of photoelectron starts from the surface after incidence of photons in about $$10^{-10}$$ s.
  • Question 6
    1 / -0
    __________ is the wavelength of photon of energy $$35$$KeV.
    $$h=6.625\times 10^{-34}$$J-s, $$c=3\times 10^8$$m/s, $$1$$eV$$=1.6\times 10^{-19}$$J.
    Solution
    Energy of photon  $$E = 35 \ KeV$$
    $$\therefore$$  $$E = 35\times 1000\times 1.6\times 10^{-19} \ J = 5.6\times 10^{-15} \ J$$
    Wavelength of photon   $$\lambda = \dfrac{hc}{E}$$
    $$\implies \ \lambda = \dfrac{6.625\times 10^{-34}\times 3\times 10^8}{5.6\times 10^{-15}} = 35\times 10^{-12} \ m$$
  • Question 7
    1 / -0
    The minimum energy required for the electron emission from the metal surface can be supplied to the free electrons by which of the following physical processes?
    Solution
    For electron emission energy is required. This energy can be supplied by Thermionic emission, field emission, photoelectric emission.
    Answer all of these
  • Question 8
    1 / -0
    Following graphs show the variation of stopping potential corresponding to the frequency of incident radiation (F) for a given metal. The correct variation is shown in graph ( $$v_0$$ = Threshold frequency) :

    Solution
    The stopping potential is the potential necessary to stop any electron from reaching outside. When the frequency of radiation is beyond these threshold frequency electrons are emitted from the metal surface. Threshold frequency is the minimum frequency require to remove an electron from the metal surface. So at $$ V_{o} $$ (threshold frequency) the electrons are removed and to stop these electrons much energy is needed ie, the stopping potential increases with increase in frequency after$$ V_{o} $$ because more electrons are emitted from the metal surface. 

    So option A is correct.
  • Question 9
    1 / -0
    If the uncertainly in the position of an electron is $$10^{-10}m$$, then the value of uncertainty in its momentum will be __________ $$kgms^{-1}$$. $$(h=6.62\times 10^{-34}J-s)$$.
    Solution
    Given :  $$h = 6.62\times 10^{-34} \ J.s$$              $$\Delta x = 10^{-10}\  m$$
    value of uncertainty in its momentum   $$\Delta p = \dfrac{h}{2\pi \Delta x}$$
    $$\implies \ \Delta p = \dfrac{6.62\times 10^{-34}}{2\times 3.142\times 10^{-10}} = 1.05\times 10^{-24} \ kg.m/s$$
  • Question 10
    1 / -0
    Though Albert Einstein did not get the Nobel Prize for his work on relatively theories, but the word 'relativity' has become synonym with Einstein. Which of the following books on relatively Einstein has written?
    I. The Theory of Relativity
    II. The Principle of Relativity
    III. The Meaning of Relativity
    IV. The Special Theory of Relativity
    V. Relativity: Special and General Theory
    The correct combination is?
    Solution

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