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Dual Nature of Radiation and Matter Test - 19

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Dual Nature of Radiation and Matter Test - 19
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  • Question 1
    1 / -0
    The frequency of radio waves corresponding to a wavelength of $$10\ m$$ is
    Solution
    Wavelength, $$\lambda = 10 \: m$$

    Using the relation $$\nu=\dfrac{c}{ \lambda }= \dfrac{ 3 \times 10^8}{10}= 3 \times 10^7\: Hz$$
  • Question 2
    1 / -0
    The de-Broglie wavelength $$\lambda$$ of a particle 
    Solution
    De-Broglei wavelength, $$\lambda  = \dfrac{h}{p}$$ where, $$h$$ is Planck's constant
     Impulse, $$I  = \Delta p    =  p$$                                
    $$\therefore$$    $$\lambda  = \dfrac{h}{I}$$
    $$\implies$$$$\lambda  \propto  \dfrac{1}{I}$$
    Thus, De-Broglei wavelength is inversely proportional to the impulse.
    Also, $$p = mv$$
    $$\therefore$$   $$\lambda = \dfrac{h}{mv} $$$$\implies  \lambda  \propto  \dfrac{1}{m}$$
  • Question 3
    1 / -0
    For the Bohr's first orbit of circumference $$2\pi r$$, the de-Broglie wavelength of revolcing electron will be.
    Solution
    $$mvr=\displaystyle \frac{nh}{2\pi}$$, according to Bohr's theory
    $$\Rightarrow 2\pi r=n\left(\displaystyle\frac{h}{mv}\right)=n\lambda$$ for $$n=1, \lambda =2\pi r$$
  • Question 4
    1 / -0
    The de-Broglie wavelength of electron in second Bohr's orbit is equal to ____________.
    Solution
    Conservation of angular momentum gives:
    $$mvr=\dfrac { nh }{ 2\pi  } \\ \Rightarrow \dfrac { h }{ mv } =\dfrac { 2\pi r }{ n }$$
    De Broglie's wavelength, $$\lambda =\dfrac { h }{ mv }$$
     Thus, for $$n =2$$
    $$\Rightarrow \lambda =\dfrac { 2\pi r }{ 2 } =\dfrac { circumference }{ 2 } $$
  • Question 5
    1 / -0
    The de-Broglie wavelength of an electron in the ground state of the hydrogen atom is :
    Solution
    Bohr's postulate for angular momentum states, 
    $$mvr=\dfrac{nh}{2\pi}$$

    For ground state, momentum of electron$$=p=mv=\dfrac{h}{2\pi r}$$

    Thus de-broglie wavelength for the electron is $$\dfrac{h}{p}=2\pi r$$
  • Question 6
    1 / -0
    A particle of mass M at rest decays into two masses $$\displaystyle { m }_{ 1 }$$ and $$\displaystyle { m }_{ 2 }$$ with non zero velocities. The ratio of de-Broglie wavelengths of the particles $$\displaystyle \frac { { \lambda  }_{ 1 } }{ { \lambda  }_{ 2 } } $$ is:
    Solution
    Initial momentum of the particle is zero as it is at rest.
    Thus according to law of conservation of momentum, the final momentum of the system must be zero.
    Let the momentum of the two fragments be $$p_1$$ and $$p_2$$.
    $$\implies$$   $$p_1  = p_2$$ 
    de-Broglie wavelength       $$\lambda = \dfrac{h}{p}$$
    where  $$p$$ is the momentum of the particle and $$h$$ is Planck's constant.
    $$\therefore$$   $$\lambda_1  = \dfrac{h}{p_1}$$      and        $$\lambda_2 = \dfrac{h}{p_2}$$
    $$p_1  = p_2$$        $$\implies$$   $$\lambda_1  = \lambda_2$$
  • Question 7
    1 / -0
    According to de Broglie, Which of the following statements is true about the wavelength of a moving particle?
    Solution
    de Broglie wavelength associated with the particle          $$\lambda = \dfrac{h}{p}$$
    $$\implies$$        $$\lambda \propto \dfrac{1}{p}$$                 where  $$h$$ is the Planck's constant
    Thus de Broglie wavelength is inversely proportional to the momentum of the particle
  • Question 8
    1 / -0
    Which of the following is the best definition of the uncertainty principle?
    Solution
    According to Heisenberg uncertainty principle, the position and velocity (hence, momentum) cannot both be measured exactly, at the same time.
  • Question 9
    1 / -0
    Einstein's theory of relativity is based on:
    Solution
    The main postulate of Einstein's special theory of relativity is that the speed of light is constant and equal to  $$c$$  in all the inertial frames  i.e the velocity of light is same for all observers. 
  • Question 10
    1 / -0
    According to Einstein, the mass of an object increases as its speed increases. This increase in mass comes from :
    Solution
    Relativistic mass of a moving object         $$m =\dfrac{m_o}{\sqrt{1 - v^2/c^2}}$$            where  $$m_o$$ is the rest mass
    $$\implies$$   $$m> m_o$$   for  $$v>0$$. Hence the mass of moving object increases.
    Using Einstein's mass-energy equivalence,     $$E = mc^2$$
    This states that the energy can be converted into mass and vice versa.
    Hence the kinetic energy gets converted into the mass.
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