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Dual Nature of Radiation and Matter Test - 48

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Dual Nature of Radiation and Matter Test - 48
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  • Question 1
    1 / -0
    A particle of mass M at rest decays into particles of masses $$m_{1} \ and  \ m_{2}$$ having non -zero velocities. The ratio of the de broglie wavelengths of the particles $$\lambda _{1}/\lambda _{2}$$ is :
    Solution
    When mass $$M$$ at rest (initial momentum$$=0$$) decays to smaller masses $${m}_{1}$$ and $${m}_{2}$$, energies  have same momentum (From conservation of momentum)
    $${ \vec { P }  }_{ 1 }+{ \vec { P }  }_{ 2 }=0\Rightarrow { P }_{ 1 }={ P }_{ 2 }$$ (magnitude)
    And de-Broglie wavelength
    $${ \lambda  }_{ 1 }=\cfrac { h }{ { P }_{ 1 } } ;{ \lambda  }_{ 2 }=\cfrac { h }{ { P }_{ 2 } } $$ (H-Plancks constant)
    Thus we get
    $${ \lambda  }_{ 1 }={ \lambda  }_{ 2 }\Rightarrow \cfrac { { \lambda  }_{ 1 } }{ { \lambda  }_{ 2 } } =1$$
  • Question 2
    1 / -0
    De-Broglie wavelength associated with thermal neutrons at room temperature of $$27^{0}$$C is :
    Solution

    Given,

    $$T=27+273=300K$$

    $${ \text { Boltzmann's constant. } }$$

    $${ k = 1.38 \times 10 ^ { - 23 } J m o l ^ { - 1 } k ^ { - 1 } }$$

    Kinetic energy of neutron at absolute temperature

    $$\text{ T is given by, }E=\dfrac{3}{2}kT$$

    $${ \lambda = \dfrac { h } { \sqrt { 2 m E } } }$$$$=\dfrac{h}{\sqrt{3mkT}}$$

    $${ \lambda = \dfrac { 6.63 \times 10 ^ { - 34 } } { \sqrt { 3 \times 1.675 \times 10 ^ { - 27 } \times 1.38 \times 10 ^ { - 23 } \times 300 } } }$$$$=1.45\times {{10}^{-10}}m\text{ }$$

    Hence, De Broglie wavelength is $$1.45\times {{10}^{-10}}m\text{ }$$

  • Question 3
    1 / -0
    The ratio of wavelength of deutron and proton accelerated through the same potential difference will be-
    Solution

  • Question 4
    1 / -0
    The minimum frequency v of continuous X-rays is related to the applied potential difference V as?
    Solution
    As we know,
    $$E=hv=eV $$
    $$\Rightarrow v\propto V$$
  • Question 5
    1 / -0
    If the mass of neutron $$= 1.7 \times 10 ^ { - 27} \mathrm { kg }$$ . Then the de broglie wavelength of neutron of energy3$$\mathrm { eV }$$ is
    Solution

  • Question 6
    1 / -0
    If Bohr radius is $$r_{0},$$ the corresponding de Broglie wavelength of the electron is 
    Solution

    Given that,

    Radius $$r={{r}_{0}}$$

    We know that,

    $$2\pi r=n\lambda $$

    Suppose $$n=1$$

    Then,

    $$\lambda =2\pi {{r}_{0}}$$

    Hence, this is the required solution

  • Question 7
    1 / -0
    A microscope using suitable photons is employed to locate an electron in an atom within a distance of $$0.1 \ \mathring {A}$$. The uncertainty of its velocity is 
    Solution
    According to Heisenberg's Uncertainty principle,
    $$m=9.11\times { 10 }^{ -31 }kg\quad \quad h=6.626\times { 10 }^{ -34 }Js$$
    $$\Delta v=\dfrac { h }{ 4\pi m\Delta x } $$
    $$\Delta v=\dfrac { 6.626\times { 10 }^{ -34 } }{ 4\times 3.14\times 9.11\times { 10 }^{ -31 }\times 0.1\times { 10 }^{ -10 } } $$
    $$\Delta v=5.79\times { 10 }^{ 6 }m/s$$
  • Question 8
    1 / -0
    What is the maximum wavelength of the Lyman series of $$He^+$$ ion ?
  • Question 9
    1 / -0
    Photons of wavelength $$6620A^0 $$ are incident normally on a perfectly reflecting screen. Calculate the number of photons paper per second falling on the screen as the total power of photons such that the exerted force is 1N.
    Solution
    Momentum, $$p=\dfrac { h }{ \lambda  } $$ reflected normally 
    $$\Delta p=\dfrac { 2h }{ \lambda  } $$
    Force $$=$$ rate of charge of momentum
    $$1=\frac { \Delta p }{ t } $$ for each photon $$(t=1sec)$$
    let $$n$$ be no. of photon
    $$1=\dfrac { n\times 2\left( 6.63\times { 10 }^{ -34 } \right)  }{ 6620\times { 10 }^{ -10 } } $$
    $$n=\dfrac { 6620\times { 10 }^{ -10 } }{ 2\times 6.63\times { 10 }^{ -34 } } $$
        $$\approx 5\times { 10 }^{ 26 }$$ photons.
  • Question 10
    1 / -0
    When an electron de-excited back from $$(n+1)^th$$ state to $$n^th$$ state in a hydrogen like atoms, wavelength of radiation emitted is $$\lambda_{1}(n>>1)$$. In the same atom de-broglies wavelength associated with an electron in $$n^th$$ state is $$\lambda_{2}$$. Then $$\dfrac{\lambda_{1}}{\lambda_{2}}$$ is proportional to
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