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Dual Nature of Matter and Radiation MCQs for NEET — Physics Questions with Answers

Practice free Dual Nature of Matter and Radiation (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

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Photons of enerty 1 eV and 2.5 ev successively illuminate a metal, whose work function is 0.5 eV, the ratio of maximum speed of emitted electrion is..........................

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Explanation

$ ( {1 \over 2 mv^2 _{max} }) _1 = (hf)_1 - \phi = 1 - 0.5 $ $ ( {1 \over 2 mv^2_ {max} }) _2 = (hf)_2 - \phi = 2.5 - 0.5 $ $ \therefore {(v^2 _{max} )_1 \over (v^2 _ {max }) _2 } = { 0.5 \over 2 } = { 1 \over 4 } $ $ \therefore {(v^2 _{max} )_1 \over (v^2 _ {max }) _2 } = { 1 \over 2 } = 1:2 $

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A proton and an x-particle are passed through same potential difference. If their initial velocity is zero, the ratio of their de Broglie's wavelength after getting accelerated is...................

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Explanation

$ m_p = 1 q_p = e $ $ \lambda = { h \over \sqrt {2meV}} $ $ \therefore \lambda \alpha { 1 \over \sqrt {mq} } $ $ \therefore { \lambda_p \over \lambda_\alpha } = \sqrt { m_\infty q _ \infty \over m_p q _p }$ $ = \sqrt { 4 \times 2e \over 1 \times e } $ $ = \sqrt 8 $ $ = 2 \sqrt 2 $ $ \therefore { \lambda _p \over \lambda_\alpha } = 2 \sqrt {2} : 1 $

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The uncertainty in position of a particle is same as its de-Broglie wavelength, uncertainty in its momentum is.......................

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Explanation

$ \triangle x .\triangle p = { h \over 2 \pi } = h $ $ \triangle x = \lambda $ $ \triangle = {h \over \lambda} $

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Wavelength of an electron having energy 10 keV is……….. $ A ^ \circ $

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Explanation

$ \lambda = { h \over \sqrt {2mE} } = { 6.6 \times 10^{-34} \over \sqrt { 2 \times 9.1 \times 10^{-31} \times 10^4 \times 1.6 \times 10^{-19} }}$ $ = 0.12 \times 10^{-10} m $ $ \lambda = 0.12 A ^\circ $

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The de-Broglie wave length of a particle having velosityof 2.25 $108 ms^ {–1 } $ , is the same value of a photon wavelength, then the ratio of kinetic energy and photon energy of the particle is.…..(take c = $ 3 \times 10^8 ms^{-1} $)

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Explanation

$ { K.E of practical \over K.E of photon} = { {1 \over 2 } mv^2 \over hf } = { {1 \over 2 } mv.v \over {hc \over \lambda} } = { {1 \over 2 } Pv \lambda \over hc }$ $ = { {1 \over 2 } {h \over \lambda } \lambda v \over hc } = { v \over 2c} ={ 2.25 \times 10^8 \over 2 \times 3 \times 10^8 } = {3 \over 8 } $

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The ration of de - Begli wavelengths of molecules of hydrogen and helium which are at temperature $ 27 ^ \circ Cand 127 ^ \circ C $ respectively is...................

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Explanation

$ { 1 \over 2 } mv_m^2 = { hc \over \lambda} - \phi $ $ = 2 - \phi (eV) ....(1) $ $ {1 \over 2 } m 4 v_m^2 = { hc \over {3 \over 4 } \lambda } - \phi $ $ = { 4 \over 3 } \times 2 - \phi = {8 \over 3 } -\phi $ $ \therefore 4 ( { 1 \over 2 } mv_m^2 ) = { 8 \over 3 } - \phi $

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A photon, an electron and a uranium nucleus all have same wavelength. The one with the most energy..................

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A light of frequency 1.5 times the threshold frequency,is incident on photo-sensitive material. If the frequency is halved and intensity is doubled, the photoelectric current becomes....................

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Explanation

final friquency $ { 1.5 \over 2 } f_0 = 0.75 f_0 < f_0 $ therefore photoelectry current is zero

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  1. The cathode of a photoelectric cell is changed such that the work function changes from $W_1$ . If the currents before and after change are $ I_1 $ and $I_2$ , all other conditions remaining unchanged, then assuming$ hf > W _2 $
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The work function of a metal is 1 eV. Light of wavelength $ 3000 A ^ \circ $ is incident on this metalsurface. The maximum velocity of emitted photoelection will be..................

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Explanation

$ { 1 \over 2 } mv^2 max = { hc \over \lambda } - \phi $ $ = ( {6.62 \times 10^{-34} \times 3.0 \times 10^8 \over 3.0 \times 10^{-7} \times 1.6 \times 10^{-19}} -1) eV $ $ = 3.1375 eV = 5.02 \times 10 h -19 J $ $ \therefore V_{max} = \sqrt { 2 \times 5.0 \times 2 \times 10^{-19} \over 9.1 \times 10^{-31 } }$ $ ( \therefore m = 9.1 \times 10^{-31} kg )$ $ = 1.05 \times 10^6 = 1 \times 10^6 ms^{-1} J$

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