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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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How many photons of red coloured light having wavelength $ 8000 A ^ \circ $ will have same energyas one photon of violet coloured light of wavelength $ 4000 A ^ \circ $ ?

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Explanation

$ E_1 = {hc \over \lambda_1} $ $ E_2 ={ nhc \over \lambda_2} $ $ E_1 = E_2 $ $ \therefore {hc \over \lambda_1} = {nhc \over \lambda_2} $ $ \therefore n = {\lambda_1 \over \lambda_2} = {8000 \over 4000} = 2 $

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Output power of He-Ne LASER of low energy is 1.00 mW. Wavelength of the ligth is 632.8 nm. What will be the number of photons emitted per second from this LASER ?

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Explanation

$ = 1.0 mW = 10^{-3} W $ $ \lambda = 632.8 nm = 632 \times 10^{-9} m $ $\lambda = 632.8 nm = 632 \times 10^{-9} m $ $ P = { nhc \over \lambda } $ $ \therefore n = { p \lambda \over hc } $ $ = { 10^{-3} \times 6.3328 \times 10^{-7} \over 6.625 \times 10^{-34} \times 3 \times 10^ {3}} $ $ ={ 6.328 \times 10^{-20} \over 19.875 \times 10^{26} } $ $=0.318 \times 10^{16} $ $\therefore n = 318 \times 10^{15} s^{-1} $

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A star which can be seen withnaked eye from Earthhas intensity $ 1.6 \times 10^{-9} Wm^{-2} $ on Earth. If the corresponding wavelength is 560 nm, and the diameter of the human eye is$ 2.5 \times 10^{-3} m $ , the number of photons entering in our in 1 s is..............

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Explanation

$ 1 = 1.6 \times 10^ {-19} w/m^2 $ $ \lambda = 560 nm = 5.6 \times 10^ {-7} m $ $ r = 2.5 \times 10^{-3} m $ $ t = ls , n = ? $ $ 1 = { E \over At} = { P \over A } $ $ \therefore P = 1A ^\circ = 1( \pi r^2 ) = 1.6 \times 10^{-9} \times 3.14 \times 6.25 \times 10^ {-6} $ $ 31.4 \times 10^{-15} W $ $ \therefore = P = { nhc \over \lambda }$ $ \therefore = { p \lambda \over hc } = { 31.4 \times 10^ { -15 } \times 5.6 \times 10^{-7} \over 6.62 \times 10^{-34} \times 3 \times 10^3 }$ $ \therefore n = 8.85 \times 10^4 $

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What should be the ratio of de-Broglie wavelength of an atom of nitrogen gas at 300 K and 1000 K. Mass of nitrogen atom is $4.7 \times 10^{-26}$ kg and it is at 1 atm pressure Consider it as an ideal gas

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Explanation

$ T _1 = 300 K $ $ T_2 = 1000 K $ $ m = 4.7 \times 10^{-26} kg $ $ P = 1 atm $ $ { 1 \over 2 } mv ^2 = { 3 \over 2 } KT $ $ \therefore m^2 v^2 = p^2 = 3 mKT$ $ \therefore p = \sqrt {3mKT} $ $ \lambda = { h \over p } $ $ \therefore \lambda = { h \over \sqrt { 3mKT} }$ $ \therefore \lambda \alpha { 1 \over \sqrt T} $ $ \therefore { \lambda _1 \over \lambda_2 } = { \sqrt { T_2 \over T_1 }} = \sqrt { 1000 \over 300 } = \sqrt {10 \over 3 } $ $ \therefore { \lambda _1 \over \lambda_2 } = 1.826$

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Wavelength of light incident on a photo - sensitive surface is reduced form $ 3500 A ^ \circ $ to 290 mm. The change in stopping potenital is....... $( h = 6.625 \times 10^{-24} J.s)$

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Explanation

$ \lambda_1 = 3500 A ^\circ , \lambda _2 =290 nm $ $ h = 6.625 \times 10^ { -24} J.s $ $ { 1 \over 2 } mv^2 max = { hc \lambda } - \phi = eV_0 $ $ \therefore V_01 e = { hc \over \lambda _1 } -\phi $ $ \therefore V_02e = { hc \over \lambda_1} -\phi $ $ \therefore V_02 e = { hc \over \lambda_2} - \phi $ $ (V_02 -V_01) e = hc ( {1\over \lambda_2 } - { 1 \over \lambda_1} )$ $ \therefore V_02 -V_01 = { hc \over e } [ { \lambda_1 - \lambda_2 \over \lambda_1 \lambda_2} ] = 12.42 = [ { 0.6 \over 3.5 \times 2.9 } ]$ $ = 0.7342$ $ = 73.42 \times 10^{-2} V $

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An electric bulb of 100 W converts 3% of electrical energyinto light energy. If the wavelength of light emitted is $ 6625 A ^ \circ $ , the number of photons emitted is 1 s is........ $( h = 6.625 \times 10^{-34} J.s)$

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Explanation

$ \lambda = 6625 A^\circ $ $ = 6.625 \times 10^ { -7} m $ $ c = 3 \times 10^8 m/s $ $ E = nhf $ $ E = { nhc \over \lambda } $ $ n = { E \lambda \over hc } = { 3 \times 6.625 \times 10^{-7} \over 6.625 \times 10^{-34} \times 3 \times 10^ 8 }$ $ \therefore n = 10^{19 } $

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Work function of Zn is 3.74 eV. If the sphere of Zn is illuminated by the X-ray of wavelength $ 12 A^ \circ $ the maximum potential produced on the sphere is ……...$( h = 6.625 \times 10^{-34} J.s)$

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Explanation

$ eV_0 = { hc \over \lambda } - \phi $ $ \therefore V_0 = { hc \over \lambda c } = { \phi \over e } $ $ \therefore V_0 = 1031.4V $

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Consider the radius of a nucleus to be $10 ^{-15}$ m . If anelectron is assumed to be in suchnucleus, what ill be its energy ? $( me = 9.1 \times 10 ^ {-31} kg ,h = 6.625 \times 10^{-34} J.s)$

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Explanation

$ \triangle x = 2 r = 2 \times 10^ {-15 } m $ $ \triangle . \triangle p = { h \over 2 \pi } $ $ \therefore \triangle p = { h \over 2 \pi \triangle x } = 0.5274 \times 10^ { -19} $ $ E = { p^2 \over 2m } P = \triangle p $ $= 9.55 \times 10^ 9 eV = 9.55 \times 10^ { 3} MeV$

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A proton falls freely under gravity of Earth. Its de Broglie wavelengthafter 10 s of its mortion is , Neglect the forces other than gravitational force. $(g = 10 {m \over s^2}, m_p = 1.6 \times 10^{-27} kg , h = 6.625 \times 10^{-34} J.s)$

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Explanation

To find the de Broglie wavelength of a proton after falling freely under gravity for 10 seconds, we first calculate the velocity (v) using the equation of motion: $$v = u + gt$$, where initial velocity (u) is 0, g is 10 m/s², and t is 10 s. Thus, $$v = 0 + (10 m/s² imes 10 s) = 100 m/s$$. The momentum (p) of the proton is given by: $$p = mv$$, where m is the mass of the proton, $$p = 1.6 imes 10^{-27} kg imes 100 m/s = 1.6 imes 10^{-25} kg imes m/s$$. The de Broglie wavelength λ is: $$ ext{λ} = rac{h}{p} = rac{6.625 imes 10^{-34} Js}{1.6 imes 10^{-25} kg imes m/s} ext{λ} = 41.40625 imes 10^{-10} m = 41.40625 Å$$. This is closest to 39.6 Å, so the correct option is $$39.6 Å$$.

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Compare energy of a photon of X-rays having 1A wavelength withthe energy of an electron having same de Broglie wavelength $( h = 6.625 \times 10^{-34} J,s.c = 3 \times 10^8 ms^{-1} , lev = 1.6 \times 10^{-19}J)$

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Explanation

$ E_p = { hc \over \lambda} $ $ E_p = 19.87 \times 10^{-16} J $ $ E_0 = { p^2 \over 2m } = { h^2 \over \lambda^2 (2m) } $ $ \therefore E_0 = 2.41 \times 10^{-17 } J $ $ \therefore { E_p \over E_0 } = { 19.87 \times 10^{-16} \over 2.41 \times 10^ {-17 } } $ $ \therefore { E_p \over E_0} = 82.4 $

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