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Physics MCQs for NEET — Practice Questions with Answers

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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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An electron is at a distance of 10 mform a charge of 10 C. Its total energy is $15.6 \times 10^{-10}$ . Its de Broglie wavelength at this point is $( h = 6.625 \times 10^{-34} J,me = 9.1 \times 10 ^ {-31} kg , K = 9 \times 10^9 SI)$

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Wavelength of an electron having energy E is where m is the mass of an electron.Find the wavelength of the electron when it centers in X-direction in the region having potential $X- V_(x)$ If we imaging that due to the potential, electron enters from one medium to another, what is the refractive index of the medium ?

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Explanation

$E = K + U $ $ E = { p ^ 2 \over 2m } - eV_{(x)} $ $ \therefore p = ( 2mE + eV_{(x)} )^{1/2} $ $ \therefore \lambda = { h \over p } = { h \over [ 2m ( E + eV_{(x)} )]^{1/2} }$ $ \mu = { \lambda_0 \over \lambda } = { h \over \sqrt { 2mE} } \times { [2m ( E + eV_{(x)})]^{1/2} \over h }$ $ \mu = [ { E + eV_{(x)} \over E} ] ^ { 1 /2} $

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U. V. light of wavelength 200 mm is incident on polished surface of Fe. work function of the surface is 4.5 eV. Find maximum speed of photo electrons $ ( h = 6.625 \times 10^{-34} J.s , c = 3 \times 10^ 8 ms^{-1} , 1 eV = 1.6 \times 10^{-19} J ) $

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Explanation

$ {1 \over 2 } mv^2 max = eV_0 = { hc \over \theta } - \phi $ $ = { 6.625 \times 10^ {-34} \times 3 \times 10^8 \over 2 \times 10^{-7} }$ $ = 4.5 \times 1.6 \times 10^ {-19} $ $ \therefore V_0 = { 9.94 \times 10^{-19} \over 1.6 \times 10^{-19} } = { 4.5 \times 1.6 \times 10^{-19 } \over 1.6 \times {-19 } } $ $ = 6.21 -4.5 $ $ \therefore V_0 = 1.71 V $ $ { 1\over 2} mv^2 max = eV_0 = 1.71 \times 1.6 \times 10^{-19} $ $ \therefore v^2 max = { 2 \times 2.74 \times 10^{-19} \over m } $ $ \therefore V_{max} = \sqrt { 5.48 \times 10^{-19} \over 9.11 \times 10^{-31} $ $ \therefore V_{max} = 7.75 \times 10^5 m/s $

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Light of $ 4560 A ^ \circ $ 1mW is incident on photo-sensitive surface of Cs (Cesium). If the quantum efficiency of the surface is 0.5% what is the amount of photoelectric current produced ?

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Explanation

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Work function of metal is 2 eV. Light of intensity $ 10^{-5} Wm ^{-2} $ is incident $ 2 cm ^ 2 $on area of it. If $ 10^{17} $ electrons of these metals absorb the light, in how much time does the photo electric effectc start ? Consider the waveform of incident light.

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Explanation

To find the time required for the photoelectric effect to start, we need to calculate the energy absorbed by the electrons and then compare it with the work function of the metal. The intensity of incident light is given as $10^{-5} ext{ Wm}^{-2}$, and the area of the metal is $2 ext{ cm}^2$ or $2 imes 10^{-4} ext{ m}^2$.

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Radius of a beam of radiation of wavelength 5000 A is $ 10^ { -3} m$ . Power of the beam is $ 10^ { -3}W $ This beam is normally incident on a metal of work function 1.9 eV. The charge emitted by the metal per unit area in unit time is...............Assume that each incident photon emits one electron. $ ( h = 6.625 \times 10 ^ { -34} J.s )$

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$ 11 \times 10^ {11} $ Photons are incident on a surface in 10 s. These photons correspond to a wavelength of $ 10 A ^ \circ $ . If the surface area of the given surface is $0.01 m^2$, the intensity of given radiations $ ( h = 6.625 \times 10 ^ { -34} J.s , c = 3 \times 10^ 8 m/s )$

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Explanation

$ t -10 s ,N = 11 \times 10^{11}, \lambda = 10A , A = 0.01 m^2 $ $ 1 = { E \over A.t} $ $ E = { NHc \over \lambda } $ $ \therefore I = { NHc \over \lambda At } $ $ \therefore I = { 11 \times 10^ {11} \times 6.625 \times 10^ {-34} \times 3 \times 10^{8} \over 10 \times 10^ {-10 } \times 10^{-2} \times 10 } $ $ = 218.6 \times 10^{-5} $ $ = 2.186 \times 10^{-3} w/m^2 $

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