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Dual Nature of Radiation and Matter Test - 26

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Dual Nature of Radiation and Matter Test - 26
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  • Question 1
    1 / -0
    A monochromatic source of light operating at 200 W emits $$4 \times 10^{20}$$ photons/second. Then the wavelength of light used is
    Solution
    $$\ P=\ \dfrac{nhc}{\lambda}$$

    $$\ 200=\ \dfrac{4\times 10^{20}\times 20\times 10^{-26}}{\lambda }$$

    $$\lambda =\ 4000\ A^{0}$$
    So, the answer is option (C).
  • Question 2
    1 / -0
    The de-Broglie wavelength of a bus moving with speed 'v' is $$'\lambda'$$ .Some passengers left the bus at a stoppage . Now when the bus moves with twice its initial speed, its kinetic energy is found to be twice its initial value. The de-Broglie wavelength now is
    Solution
    Initially
    $$\lambda =\dfrac{h}{mv}$$

    Now
    $$\dfrac{1}{2}M(2v)^{2}=2\times \dfrac{1}{2}mv^{2}$$

    $$2Mv^{2}= mv^{2}$$

    $$M=\dfrac{m}{2}$$

    $$so\ \lambda _{Now}=\dfrac{h}{\dfrac{m}{2}\times 2v}$$
    $$=\lambda$$
    So, the answer is option (A).
  • Question 3
    1 / -0
    If the momentum of an electron is changed by $$p_{m}$$, then the de Broglie wavelength associated with it increased by $$0.5\%$$. The initial momentum of electron will be
    Solution
    $$\lambda =\dfrac{h}{p}$$

    $$\lambda_i=\dfrac{h}{p_{i}}$$
    Now
    $$\lambda _{i}+\dfrac{0.5}{100}\lambda _{i}=\dfrac{h}{p-p_{m}}$$

    $$\dfrac{201}{200}(\dfrac{h}{p})=\dfrac{h}{p-p_{m}}$$

    $$201p-201p_{m}=200p$$
    $$p=201p_{m}$$
    So, the answer is option (C).
  • Question 4
    1 / -0
    The momentum of a photon of a electromagnetic radiation is $$3.3\times 10^{-29}kg \ m \ s^{-1}$$ .The frequency of the associated waves is $$(h = 6.6 \times 10^{-34} Js, c = 3 \times 10^{8} m/s)$$ 
    Solution
    $$p=3.3\times 10^{-29}\ kgms^{-1}$$

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

    $$\dfrac{c}{v}=\dfrac{h}{p}$$

    $$v=\dfrac{pc}{h}$$

    $$=\dfrac{3.3\times 10^{-29}\times 3\times 10^{8}}{6.6\times 10^{-34}}$$

    $$=1.5\times 10^{13}Hz$$
    So, the answer is option (D).
  • Question 5
    1 / -0
    Light of wavelength $$5000 A^o$$ falls on a sensitive surface. If the surface has received $$10^{-7}J$$ of energy, then the number of photons incident on the surface is nearly (given $$h= 6.6 \times 10^{-34}$$ Js $$c = 3 \times 10^{8}$$ m/s)
    Solution

     Energy= nhcλ
    Number of photons   $$N = \dfrac{E}{\dfrac{hc}{\lambda}} = \dfrac{E\lambda}{hc}$$
    where  $$E = 10^{-7} \ J$$  and  $$\lambda = 5000\times 10^{-10} \ m$$
    $$\therefore$$  $$N = \dfrac{10^{-7}\times 5000\times 10^{-10}}{(6.6\times 10^{-34})(3\times 10^8)} =2.5\times 10^{11}$$
    So, the answer is option (A).
  • Question 6
    1 / -0
    A positron and a proton are accelerated by the same accelerating potential. Then the ratio of the associated wavelengths of the positron and the proton will be :
    [ $$M$$ = Mass of proton, $$m$$ = Mass of positron]
    Solution

  • Question 7
    1 / -0
    Electrons are accelerated through a p.d. of 150V. Given $$ m=9.1\times{10}^{-31} $$ Js, the de Broglie wavelength associated with it is 
    Solution
    De-Brogtie wavelength
    $$\lambda =\dfrac{h}{mv}$$

    And
    $$\dfrac{1}{2}mv^{2}=eV$$

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

    $$\lambda =\dfrac{h}{mv}$$

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

    $$\lambda =\dfrac{6.62\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 150}}$$

    $$\lambda =1.0\ A^{0}$$
    So, the answer is option (B).
  • Question 8
    1 / -0
    The de Broglie wavelength associated with an electron of energy 500 eV is given by
    (take $$h=6.63\times 10^{-34}Js,m=9.11\times 10^{-31}kg$$ )
    Solution
    $$\lambda =\sqrt{\dfrac{150}{V}}\ A^{\circ }$$

        $$=\sqrt{\dfrac{150}{500}}\ A^{\circ }$$

        $$=0.55\ A^{\circ }$$
    So, the answer is option (D).
  • Question 9
    1 / -0
    The wavelength of de broglie waves associated with a beam of protons of kinetic energy $$5 \times 10^{2}$$eV.
    (Mass of each photon$$= 1.67 \times 10^{-27}$$Kg, $$h=6.62 \times 10^{-34}$$Js.)
    Solution
    $$\dfrac{1}{2}mv^{2}=E$$

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

    $$=\sqrt{2\times 5\times 10^{2}\times 1.6\times 10^{-19}\times 1.6\times 10^{-27}}$$

    $$=51.69\times 10^{23}\ N$$

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

    $$=\dfrac{6.62\times 10^{-34}}{51.63\times 10^{-23}}$$

    $$=1.28\times 10^{-12}\ m$$
    So, the answer is option (D).
  • Question 10
    1 / -0
    The momentum ( in Kg-m/s) of an electron having wavelength 2$$A^{0}$$  $$(h=6.62 \times10^{-34}Js.)$$
    Solution
    The relationship between de Broglie wavelength $$(\lambda)$$ and the momentum (p) of a photon is given by:
    $$\lambda=\dfrac{h}{p}$$.
    Using this, the momentum of an electron having wavelength $$2 \mathring A $$ is $$3.3125  \times 10^{-24}$$kg-m/s.
    So, the answer is option (A).
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