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Dual Nature of Radiation and Matter Test - 46

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Dual Nature of Radiation and Matter Test - 46
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  • Question 1
    1 / -0
    A cesium photo cell, with a steady potential difference of $$60$$ volt across it, is illuminated by a small bright light placed $$50$$ cm away. When the same light is placed one meter away, the photoelectrons emerging from the photo cell : (assume that potential difference applied is sufficient to produce saturation current)
    Solution

    number $$\int {photo} $$ electronic $$ \propto {1 \over {{d^2}}}$$

                    $$\displaystyle {{{N_1}} \over {{N_2}}} = {\left( {{{{d_2}} \over {{d_1}}}} \right)^2}$$

                    $$ = \displaystyle {\left( {{{100} \over {50}}} \right)^2}$$

                    $$ = \displaystyle {4 \over 1}$$

                    $${N_2} = \displaystyle {{{N_1}} \over 4}$$

  • Question 2
    1 / -0
    A free particle with initial kinetic energy $$E$$ and de-broglie wavelength $$\lambda$$ enters a region in which it has potential energy $$U$$. What is the particle's new de-Broglie wavelength?
    Solution

  • Question 3
    1 / -0
    The De-Broglie wavelength associated with electrons revolving round the nucleus in a hydrogen atom in ground state, will be-
    Solution
    According to $$Bohr's$$ postulate $$mvr=\dfrac{nh}{2\pi}$$ so the momentum in ground state$$(n=1)$$ will be $$p=mv=\dfrac{h}{2\pi r}$$
    Required wavelength will be $$\lambda= \dfrac{h}{p}=\dfrac{h}{h/2 \pi r}=2\pi r$$
    Value of radius for ground state in hydrogen is $$r=0.53Angstrom$$ 
    Using that we get $$\lambda=3.32Angstrom$$
    Option B is correct.
  • Question 4
    1 / -0
    The ratio of $$d-$$Broglie wavelength of molecular of hydrogen and oxygen kept in two vessels separately at $$27^oC$$ and $$127^oC$$ respectively is:
    Solution
    We know that,
    $$\dfrac { { \lambda H }_{ 2 } }{ { \lambda H }_{ e } } =\dfrac { \sqrt { { M }_{ { H }_{ e } }{ T }_{ { H }_{ e } } }  }{ \sqrt { { M }_{ { H }_{ 2 } }{ T }_{ { H }_{ 2 } } }  } $$
    Now, 
    $$\dfrac { { \lambda H }_{ 2 } }{ { \lambda H }_{ e } } =\sqrt { \dfrac { 4\left( 127+273 \right)  }{ 2\left( 27+273 \right)  }  } $$
    $$\dfrac { { \lambda H }_{ 2 } }{ { \lambda H }_{ e } } =\sqrt { \dfrac { 1600 }{ 600 }  } $$
    $$\dfrac { { \lambda H }_{ 2 } }{ { \lambda H }_{ e } } =\sqrt { \dfrac { 8 }{ 3 }  } =\sqrt { \dfrac { 4 }{ 3 }  } $$
  • Question 5
    1 / -0
    Which of the following figures represent the variation of particle momentum and the associated de-Broglie wavelength?
    Solution
    Hint:- According to de-Broglie hypothesis a wave is associated with a moving particle whose wavelength is given by
    $$\lambda =\dfrac {h}{P}$$

    Explanation:-
    As wavelength is given by 
    $$\lambda =\dfrac {h}{P}$$
    $$\lambda \propto \dfrac{1}{P}$$
    $$P \propto \dfrac {1}{\lambda}$$
    As momentum is inversely proportional to wavelength of light so graph between momentum and wavelength is hyperbolic.




  • Question 6
    1 / -0
    A body of mass $$m$$ is dropped freely from a height $$h$$. The de-Broglie wavelength of the body as it reaches the ground is:
    Solution
    We know ,
    Velocity$$v = \sqrt {2gH} $$
    $$\begin{array}{l} \lambda =\frac { h }{ { mv } }  \\ =\frac { h }{ { m\sqrt { 2gH }  } }  \end{array}$$
    $$\therefore$$ Option $$D$$ is correct.
  • Question 7
    1 / -0
     what is the angular momentum of an electron of de$$-$$broglie wavelength $$\lambda ?$$ given $$r$$ is the radius of orbit$$.$$
    Solution
    $$mvr = \dfrac{h}{\lambda }$$
    $$r = \dfrac{{rh}}{\lambda }$$
    Hence,
    option $$(A)$$ is correct answer.
  • Question 8
    1 / -0
    According to de Broglie, wavelength of electron in second orbit is $$10^{-9}$$ meter. Then the circumference of orbit is :-
    Solution

  • Question 9
    1 / -0
    A particle is dropped from a height H. The de-Broglie wavelength of the particle as a function of height is proportional to
    Solution
    Solution :
    Velocity $$v= \sqrt{2gH}$$
    $$\lambda =\dfrac{h}{mv}$$
    $$= \dfrac{h}{m\sqrt{2gh}}$$
    $$\lambda \alpha H^{-1/2}$$

  • Question 10
    1 / -0
    Stopping potential $$(V_0)$$ frequency $$(v)$$ of radiation graph for three different emitter surfaces $$A, B$$ and $$C$$ as shown in figure, then 
    (a) threshold frequency of $$A < B <C$$ 
    (b) work function of $$A < B <C$$ 
    (c) work function of $$A = B = C$$
    (d) threshold wavelength of $$A < B <C$$ 
    (e) ratio of the slopes of $$A , B, C$$ id $$1 : 1 : 1$$

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