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NEET NEET Questions — Free MCQ Practice

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What will be velocity of light of particle having mass double than its rest mass ?

A. $ 2c \over 3 $ B. $ c \over 2 $ C. $ \sqrt 3c \over 2 $ D. $ 3c \over 4 $

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Explanation

$ m = { m_0 \over \sqrt {1 -{v^2 \over c^2 }} } = 2m_0 $ $ \therefore \sqrt {1 -{v^2 \over c^2 }}= { 1 \over 2 } $ $ {1 -{v^2 \over c^2 }} = {1 \over 4 } $ $ \therefore {v^2 \over c^2 } = { 3 \over 4 } $ $ \therefore { v \over c }= { \sqrt 3 \over 2 }$ $ \therefore v = { \sqrt3 \over 2 } e $

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If de-Broglie wavelength of electron is increased by 1 % its momentum................

A. increases by 1% B. decreases by 1% C. increased by 2% D. decreases by 2%

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Explanation

$ \lambda = { h \over p } $ $ \therefore \lambda \alpha {1 \over p} $ $ \therefore \lambda \alpha p^{-1} $ $ d \lambda \alpha -{p ^ {-2}}dp $

$ d \lambda \alpha -{ 1 \over p ^ {-2}}dp $ $ \therefore { dp \over \lambda } \times 100 = -{ dp \over p^2} \times p \times 100 $ $\therefore { dp \over \lambda } \times 100 = -1\%$ $ decreases by \%$

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With how much p.d. should an electron be accelerated, so that its de-Broglie wavelength is $ 0.4 A ^ \circ $

A. 9410 V B. 94.10 V C. 9.140 V D. 941.0V

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Explanation

$ \therefore = { h \over \sqrt { 2Vem}} $ $ v = { h^2 \over 2me \lambda ^2 } $ $ = { (6.62 \times 10^{-34})^2 \over 2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times (0.4 \times 10^{-1} )^2 }$ = 940.5 =941 V

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de-Broglie wavelength of atom at T K absolute temperature will be....................

A. $ h \over mkT $ B. $ h \over \sqrt {3mkT} $ C. $ \sqrt {2mkT} \over h $ D. $ \sqrt {2mkT} $

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Explanation

The de-Broglie wavelength of a particle is given by the formula: $$ ext{wavelength} = rac{h}{ ext{momentum}} $$ For an atom at temperature T, the momentum can be related to the thermal energy as $$ p = rac{mv}{ ext{thermal energy}} $$ The thermal energy is given by $$ rac{3kT}{2} $$ Therefore, the de-Broglie wavelength is $$ ext{wavelength} = rac{h}{ ext{momentum}} = rac{h}{ ext{mv}} rac{1}{ ext{thermal energy}} = rac{h}{ ext{mv}} rac{1}{ ext{(3kT/2)}} = rac{h}{ ext{mv}} rac{1}{ ext{(3kT/2)}} = rac{h}{ ext{sqrt}(3mkT)} $$

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Uncertainty of momentum of particle is $ 10^{-30} kg ms^{-1}$ so minimum uncertainty in its position is m.

A. $ 10^{-8} m$ B. $ 10^{-12} m$ C. $ 10^{-16} m$ D. $ 10^{-4} m$

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An electron is accelerated between two points having potential 20 V and 40 V, de- Broglic wavelength of electron is..................

A. $ 0.75 A ^\circ $ B. $ 7.5 A ^\circ $ C. $ 2.75 A ^\circ $ D. 0.75 nm

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Explanation

$ \lambda = { h \over \sqrt{2meV} } $ v = v_2 - V_1 = 40 -20 = 20v $ = { 6.62 \times 10^{-34} \over \sqrt { 2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 20}}$ $ = 0.274 \times 19^{-9} $ $ = 2.75 A $

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de - Broglic wavelength of electron in nth Bohr orbit is............

A.

$ { 2 \pi r \over n } A^\circ $

B.

$ { 2 \pi rn } A^\circ $

C.

$ { 1 \over n } A ^\circ $

D.

$ nA ^\circ $

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Explanation

According to de Broglie hypothesis, the wavelength of an electron in the nth Bohr orbit is given by: $$ ext{wavelength} = rac{2 ext{Ï€}r}{n} $$ Here, $r$ is the radius of the nth orbit and $n$ is the principal quantum number. This formula is derived from the quantization condition of angular momentum in Bohr's model of the atom, where $mvr = nrac{h}{2 ext{Ï€}}$. Therefore, the correct option is: $$rac{2 ext{Ï€}r}{n} A^ ext{âˆ˜}$$

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In photo electric effect, if threshold wave length of a metal is $ 5000 A^ \circ $ work function of this metal is....................eV. $ ( h = 6.6 \times 10^{-34} J.s , c = 3 \times 10^8 m/s , 1 eV - 1.6 \times 10^{-19} J.s ) $

A. 1.24 B. 2.48 C. 4.96 D. 3.72

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Explanation

$ \phi = { hc \over \lambda_0 e } = { 6.62 \times 10^{-34 } \times 3 \times 10^8 \over 5 \times 10^{-7} \times 1.6 \times 10^{-19} }$ = 2.48 eV

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Photo senstive surface is incident by light having frequecy 3 times its threshold frequency. In this condition, if frequency of light is made half and intensity of light is made double, magnitude of photo electric current becomes

A. Fourth B. Double C. Half D. Zero

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If ratio of threshold frequencies of two metals is 1 : 3, ratio of their work functions is.............

A. 1:3 B. 3:1 C. 4:16 D. 16:4

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Explanation

$ { \phi_1 \over \phi_2 } = { hf_01 \over hf_02} = { f_01 \over f_03} $ $ \therefore { \phi_1 \over \phi_2 } = {1 \over 3 } $

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