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Dual Nature of Radiation and Matter Test - 14

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Dual Nature of Radiation and Matter Test - 14
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  • Question 1
    1 / -0
    An electron of mass m with an initial velocity $$\overrightarrow{V}=V_0\hat{i}(V_0 > 0)$$ enters an electric field $$\overrightarrow{E}=-E_0\hat{i}$$($$E_0=$$constant $$> 0$$) at $$t=0$$. If $$\lambda_0$$ is its de-Broglie wavelength initially, then its de-Broglie wavelength at time t is?
    Solution
    R.E.F image
    The diagram will be :- 
    Initial velocity $$ = V_{0} $$
    $$ \because $$ Acceleration is constant :-
    Final velocity $$ = v = v_{0}+\dfrac{eE_{0}}{m}t $$
    (at time t)
    Now, at $$ t = 0 \Rightarrow $$ de-Broglie wavelength $$ = \lambda _{0} $$
    $$ \Rightarrow \lambda _{0} = \dfrac{h}{mv_{0}} $$
    Let at time t, de-Broglie wavelength $$ = \lambda $$
    Then, $$ \lambda = \dfrac{h}{mV} $$
    $$ = \dfrac{h}{m(V_{0}+\dfrac{eE_{0}t}{m})} $$
    $$ = \dfrac{h}{mV_{0}+eE_{0}t} $$
    $$ = \dfrac{1}{1/\lambda _{0}+eE_{0}t/h} $$
    $$ = \dfrac{\lambda .h}{h+eE_{0}\lambda _{0}t} $$
    $$ = \dfrac{\lambda _{0}}{1+\dfrac{eE_{o}\lambda _{0}t}{h}} $$
    $$ = \boxed{\lambda = \dfrac{\lambda _{0}}{1+(\dfrac{eE_{0}}{mV_{0}})t}} $$ 

  • Question 2
    1 / -0
    Electrons of mass m with de-Broglie wavelength $$\lambda$$ fall on the target in an X-rays tube. The cutoff wavelength $$(\lambda_0)$$ of the emitted X-rays is
    Solution
    Using de-Broglie equation             $$\lambda = \dfrac{h}{p}$$            where  $$ p =\sqrt{2mE}$$
    $$\implies$$    $$\lambda = \dfrac{h}{\sqrt{2mE}}$$
    Energy of the X-ray emitted      $$E = \dfrac{hc}{\lambda_o}$$
    $$\therefore$$   $$\lambda = \dfrac{h}{\sqrt{2m \times \dfrac{hc}{\lambda_o}}}$$                     $$\implies \lambda_o = \dfrac{2mc\lambda^2}{h}$$
  • Question 3
    1 / -0
    If the kinetic energy of a particle is increased by 16 times, the percentage change in the de Broglie wavelength of the particle is:
    Solution
    Let the initial de Broglie wavelength be $$\lambda_0=\dfrac{h}{mv_0}$$
    The kinetic energy is increased 16 times.
    Thus $$KE'=16KE_0$$
    $$\implies \dfrac{1}{2}mv^2=16\times \dfrac{1}{2}mv_0^2$$
    $$\implies v=4v_0$$
    Thus the new de Broglie wavelength=$$\lambda=\dfrac{h}{mv}=\dfrac{h}{m(4v_0)}=\dfrac{\lambda_0}{4}$$

    Thus the percentage change in wavelength:   $$=\dfrac{\lambda_0-\lambda}{\lambda_0}\times 100\\=75$$%
                                       
  • Question 4
    1 / -0
    De Broglie wavelength $$\lambda$$ associated with neutrons is related with absolute temperature T as:
    Solution
    De broglie wavelength $$\lambda=\dfrac{h}{mv}$$
    where $$v$$ is the velocity of the particle.
    Let $$E$$ be the kinetic energy of the neutrons.
    Thus $$E=\dfrac{1}{2}mv^2$$
    $$\implies v=\sqrt{\dfrac{2E}{m}}$$
    Therefore $$\lambda=\dfrac{h}{\sqrt{2mE}}$$
    Since energy of neutrons $$E\propto T$$
    $$\lambda\propto \dfrac{1}{\sqrt{T}}$$
  • Question 5
    1 / -0
    If we assume kinetic energy of a proton is equal to energy of the photon, the ratio of de Broglie wave length of proton to photon is proportional to :
    Solution

  • Question 6
    1 / -0
    If velocity of a particle is three times of that of electron and ratio of de Broglie wavelength of particle to that of electron is $$1.814 \times 10^{-4}$$. The particle will be :
    Solution
    de broglie wavelength, $$\lambda = \dfrac {h}{p}$$
    $$\Rightarrow \lambda = \dfrac {h}{mv}$$

    $$\Rightarrow \dfrac {\lambda_{p}}{\lambda_{e}} = \dfrac {m_{e}v_{e}}{m_{p}v_{p}}$$

    $$\Rightarrow m_{p} = \dfrac {m_{e}v_{e}}{v_{p}}\times \dfrac {\lambda_{e}}{\lambda_{p}}$$

    Here, $$m_{e} = 9.1\times 10^{-31} kg, v_{p} = 3v_{e}$$ and $$\dfrac {\lambda_{p}}{\lambda_{e}} = 1.814 \times 10^{-4}$$
    $$\Rightarrow m_{p} = \dfrac {9.1\times 10^{-31}}{1.814\times 10^{-4} \times 3} = 1.672\times 10^{-27} kg$$

    Thus, the particle is neutron.
  • Question 7
    1 / -0
    Hard X-rays for the study of fractures in bones should have a minimum wavelength of $$\displaystyle { 10 }^{ -11 }m$$. The accelerating voltage for electrons in X-ray machine should be:
    Solution
    From conservation of energy the electron kinetic energy equals the maximum photon energy (we neglect the work function $$\displaystyle \phi $$ because it is normally so small compared to $$\displaystyle { eV }_{ 0 }$$
    $$\displaystyle { eV }_{ 0 }={ hv }_{ max }$$
    $$\displaystyle { eV }_{ 0 }=\frac { hc }{ { \lambda  }_{ min } } $$
    $$\displaystyle { V }_{ 0 }=\frac { hc }{ { e\lambda  }_{ min } } $$
    $$h$$ (Plank's Constant) = $$6.625\times 10^{-34}J-s$$ , $$c$$ (Speed of light) = $$3\times 10^{8}\ $$$$m/s$$,
     $$e$$ (Charge of electron) = $$1.6\times10^{-19}C$$
    $$\displaystyle { V }_{ 0 }=\frac { 12400\times { 10 }^{ -10 } }{ { 10 }^{ -11 } } $$
    $$\displaystyle =124\ kV$$
    Hence, accelerating voltage for electrons in X-ray machine should be less than $$124\  $$$$ kV$$.
  • Question 8
    1 / -0
    The de Broglie wavelength of a neutron when its kinetic energy is $$K$$, is $$\lambda$$. What will be its wavelength when its kinetic energy is $$4K $$?
    Solution
    Kinetic energy of neutron=$$\dfrac{1}{2}mv^2=K$$
    $$\implies v=\sqrt{\dfrac{2K}{m}}$$
    De-broglie wavelength of neturon=$$\dfrac{h}{p}$$
    $$=\dfrac{h}{mv}=\dfrac{h}{\sqrt{2mK}}\propto \dfrac{1}{\sqrt{K}}$$
    Thus when kinetic energy becomes $$4K$$, the de-broglie wavelength becomes $$\dfrac{\lambda}{2}$$.
  • Question 9
    1 / -0
    The beam of light has three wavelengths $$4144\overset {\circ}{A}, 4972\overset {\circ}{A}$$ and $$6216\overset {\circ}{A}$$ with a total intensity of $$3.6\times 10^{-3}Wm^{2}$$ equally distributed amongst the three wavelengths. The beam falls normally on the area $$1\ cm^{2}$$ of a clean metallic surface of work function $$2.3\ eV$$. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in $$2s$$.
    Solution
    As we know, threshold wavelength $$(\lambda_{0}) = \dfrac {hc}{\phi}$$
    $$\Rightarrow \lambda_{0} = \dfrac {(6.63\times 10^{-34})\times 3\times 10^{8}}{2.3\times (1.6\times 10^{-19})} = 5.404 \times 10^{-7}m$$.
    $$\Rightarrow \lambda_{0} = 5404\overset {\circ}{A}$$
    Hence, wavelength $$4144\overset {\circ}{A}$$ and $$4972\overset {\circ}{A}$$ will emit electron from the metal surface.
    For each wavelength energy incident on the surface per unit time
    $$=$$ intensity of each $$\times$$ area of the surface wavelength
    $$= \dfrac {3.6\times 10^{-3}}{3}\times (1\ cm)^{2} = 1.2\times 10^{-7} joule$$
    Therefore, energy incident on the surface for each wavelengths is $$2s$$
    $$E = (1.2\times 10^{-7})\times 2 = 2.4\times 10^{-7}J$$
    Number of photons $$n_{1}$$ due to wavelength $$4144\overset {\circ}{A}$$
    $$n_{1} = \dfrac {(2.4\times 10^{-7})(4144\times 10^{-10})}{(6.63\times 10^{-34})(3\times 10^{8})} = 0.5\times 10^{12}$$
    Number of photon $$n_{2}$$ due to the wavelength $$4972\overset {\circ}{A}$$
    $$n_{2} = \dfrac {(2.4\times 10^{-7})(4972\times 10^{-10})}{(6.63\times 10^{-34})(3\times 10^{8})} = 0.572\times 10^{12}$$
    Therefore total number of photoelectrons liberated in $$2s$$,
    $$N = n_{1} + n_{2}$$
    $$= 0.5\times 10^{12} + 0.575\times 10^{12}$$
    $$= 1.075\times 10^{12}$$.
  • Question 10
    1 / -0
    The momentum of a proton is p. The corresponding wavelength is
    Solution
    $$\lambda =\dfrac{h}{mv}$$

       $$=\dfrac{h}{p}$$
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