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    NEET Physics: Dual Nature Of Matter And Radiation Important Questions, Solutions

    NEET Physics: Dual Nature Of Matter And Radiation Important Questions, Solutions

    #NEET
    Updated on 23 Jun 2022, 09:09 AM IST

    If you want to become a doctor, it is important that you get a good rank in the National Eligibility cum Entrance Test (NEET) for undergraduate (UG) admissions to medical courses. And for this, you need to master all the subjects (Physics, Chemistry and Biology) included in the NEET UG syllabus. A good grip on Physics is crucial as 25% of the questions in the NEET paper is from the subject. So, how well you prepare for Physics is crucial in deciding your rank in the NEET exam, which is conducted annually.

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    NEET Physics: Dual Nature Of Matter And Radiation Important Questions, Solutions
    Dual Nature Of Radiation And Matter Past FIve Year Papers(Image: Shutterstock)

    Knowing the important topics from each chapter will be quite useful in your preparation. Here is an analysis of the previous five years' NEET questions from the Class 12 NCERT chapter, Dual Nature Of Matter And Radiation. In the Last five years of NEET paper, a total of 9 questions were asked from the Class 12 NCERT chapter, Dual Nature Of Matter And Radiation.

    Topic Wise Distribution Of Questions

    Topic

    Number Of Questions

    Einstein’s Photoelectric Equation

    3

    de-Broglie Wavelength

    3

    Concept of Electron Volt

    1

    Energy of Photon

    1

    Intensity of Light

    1

    The above table makes it clear that Einstein's Photoelectric Equation and the de-Broglie Wavelength are two important topics of the chapter, Dual Nature Of Matter And Radiation. Let us get a brief idea of these concepts and the NEET questions related to them.

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    Einstein’s Photoelectric Equation

    The photoelectric equation is given by -

    Kinetic Energy (K.E) of the photo-electrons = (Energy obtained from the Photon) – (The energy used to escape the metallic surface).

    KE=h?-ɸ

    ɸ=h?0

    The energy used to escape the metallic surface is known as the work function (ɸ). Here ?0 is the threshold frequency.

    Q.1 NEET - 2017

    The photoelectric threshold wavelength of silver is 3250 x 10-10 m. The velocity of the electron ejected from a silver surface by the ultraviolet light of wavelength 2536 x 10-10 m is:

    (Given h = 4.14 x 10-15 eVs and c = 3 x 108 ms-1)

    Solution

    Here the wavelengths are given.

    K.E.=\frac{hc}{\lambda },\ \nu=\frac{c}{\lambda },\ \nu_0=\frac{c}{\lambda_0 }

    Therefore the photoelectric equation can be written as

    \\\frac{hc}{\lambda }=hc(\frac{1}{\lambda }-\frac{1}{\lambda _{o}})\\\frac{1}{2}mv^2=hc(\frac{1}{\lambda }-\frac{1}{\lambda _{o}})

    v=\sqrt{\frac{2hc}{m}.\left ( \frac{1}{\lambda }-\frac{1}{\lambda _{o}} \right )}

    Given

    \\ h =6.67\times 10^{-34}J-s\\ c=3\times 10^{18}m/s\\ \lambda =2536\times 10^{-10}m\\ \lambda_{0} = 3250 \times 10^{-10} \\m=9.1\times 10^{-31}kg

    Substituting the values the answer is

    \approx 6 \times 10^5 \text{ms}^{ - 1}

    Q.2 NEET - 2018

    When the light of frequency 2?0 (where ?0 is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is ?1. When the frequency of the incident radiation is increased to 5?0, the maximum velocity of electrons emitted from the same plate is ?2. The ratio of ?1 to ?2 is

    Solution

    From the Einstein equation

    1/2 mv^{2}= h{v}-\phi = h(v-v_{0})

    Initially

    1/2 mv_{1}^{2}= h(2v_{0}-v_{0})=hv_{0}

    Finally

    1/2 mv_{2}^{2}= h(5v_{0}-v_{0})=4hv_{0}

    \Rightarrow \frac{v_{2}^{2}}{v_{1}^{2}}=4 or

    \Rightarrow \frac{v_{1}}{v_{2}}=1/2

    Q.3 NEET - 2021

    An electromagnetic wave of wavelength λ is incident on a photosensitive surface of negligible work function. If 'm' mass is of photoelectron emitted from the surface has de-Broglie wavelength λd then:

    Solution

    \frac{hc}{\lambda }=\phi +kE_{max}

    Momentum P = h/λd where λd is the de-Broglie wavelength, work function ɸ=0(Given)

    \frac{hc}{\lambda }=0+\frac{P^{2}}{2m}=\frac{\left ( h/\lambda_d \right )^{2}}{2m}

    \frac{hc}{\lambda }=\frac{h^2}{\lambda ^{2}_{d}\times 2m}

    \left ( \frac{2mc}{h} \right )\lambda ^{2}_{d}=\lambda

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    The de Broglie Wavelength Of Electron

    The de-Broglie wavelength of electron λe is given by

    \lambda_e = \frac{h}{m_ev}= \frac{h}{\sqrt{2m_eK}}=\frac{h}{\sqrt{2m_e(eV)}}

    h=6.626\times 10^{-34} \ Js

    m_e=9.1\times 10^{-31} \ kg

    e=1.6\times 10^{-19} \ C

    \lambda _{e }= \frac{12.27}{\sqrt{V}}A^{\circ}

    The de Broglie Wavelength Associated With Charged Particle

    \lambda = \frac{h}{\sqrt{2mK}}= \frac{h}{\sqrt{2mqV}}

    Where K is the Kinetic Energy, m is the mass of the particle, q is the charge of the particle and V is the potential difference.

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    Q.4 NEET - 2017

    The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T (Kelvin) and mass m, is

    Solution

    Kinetic energy of the neutron in thermal equilibrium with heavy water is-

    K.E=\frac{3}{2}kT

    =>\lambda =\frac{h}{\sqrt{2mK.E}}=\frac{h}{\sqrt{2m.\left ( \frac{3}{2}kT \right )}}

    Here K.E is the Kinetic Energy

    \Rightarrow \lambda =\frac{h}{\sqrt{3mkT}}

    Q.5 NEET - 2018

    An electron of mass m with an initial velocity

    \overset{\rightarrow}V = V_{o} \hat{i} \left ( V_{o} > 0 \right )

    enters an electric field

    \overset{\rightarrow}E = -E_{o} \hat{i} \left ( E_{o} = constant > 0 \right )

    at t = 0 . If \lambda _{o} is its de-Broglie wavelength initially, then its de-Broglie wavelength at time t is

    Options

    1. \lambda_{o}t

    2. \lambda _{o}\left ( 1 + \frac{eE_{o}}{mv_{o}}t \right )

    3. \frac{\lambda _{o}}{\left ( 1 + \frac{eE_{o}}{mv_{o}}t \right )}

    4. \lambda_{o}

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    Solution

    F=-e(-E_0\hat{i})=eE_0\hat{i}

    a=eE_0\hat{i}/m

    (\text{Acceleration is uniform, so from v=u+at})

    At time t the magnitude of the velocity

    v=v_{0}+\frac{eE_{0}}{m}t

    Momentum

    p(t)=m(v_{0}+ \frac{eE_{0}t}{m})

    \\(h/mv_0=\lambda_0)\\\lambda = \frac{h}{p(t)} = \frac{h}{mv_{0}(1+ \frac{eE_{0}t}{mv_{0}})}=\frac{\lambda _{0}}{1+\frac{eE_{0}t}{mv_{0}}}\

    So option C is the right answer.

    Q.6 NEET - 2019

    An electron is accelerated through a potential difference of 10000 V . Its de Broglie wavelength is

    ( m_ e ) = 9 \times 10 ^{-31} Kg

    Solution

    Use the equation-

    \lambda _{e }= \frac{12.27}{\sqrt{V}}A^{\circ}

    \lambda _{electron}= \frac{12.27}{\sqrt{10^4}}A^{\circ}=12.27*10^{-12}m

    In the last 5 years’ (2017-2021) NEET papers, 9 questions were asked and out of these 6 were from the topics de Broglie Wavelength and Einstein's Photoelectric Equation. That is, 66.67% of questions from Dual Nature were from these two topics.

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