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Physics NEET Questions — Free MCQ Practice

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In terms of Rydergi constant R. The wave number of first Balmer line is

A. $ 5R \over 36 $ B. $ 8R \over 9$ C. R D. $ 8R \over 20 $

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The hydrogen atom can give spectral lines in the series Lyman, Balmer and Paschen.which of the following statement is correct

A. Lyman series is in the infrared region B. Balmer series is in the ultraviolet region C. Balmer series is in the visible region D. Paschen series is in the visible region.

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Explanation

$ wave number = {1 \over \lambda} = R [{1 \over 2^2} - {1 \over 3^2}] = R[{1\over 4} - {1 \over 9}] = { 5R \over 36 }$

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A gamma ray Photon creates an electron- Position Pair. If the rest mass energy of an electron is 0.5 MeV. and the total kinetic energy 0.7 MeV, then the energy of the gamma ray Photon must be

A. 3.9 MeV B. 1.78 MeV C. 0.78 eV D. 0.16 eV

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Explanation

$Ionigation energy of atom E = {13.6 Z^2 \over n^2} ev$ $For helium z = 2= { 13.6 \times (2)^2 \over (1)^1 } = 54.4 ev$

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Large angle scattering of $\alpha$ - particle could not be explained by

A. Thomson modal B. Ruther Rutherford and Thomson modal C. Both rutherford and Thomson modal D. Neither Ruthorford modal nor Thomson modal

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Explanation

$mass \, defect \, \triangle m = ( 2mp +2mn) -MHe$ $ \triangle m = ( 2 \times 1.0087 -2 \times 1.0073) - 4.0015 $ = 4.032-4.0015 $ \triangle m = 0.030 amu$ $ E = \triangle m \times 931.48 mev$ $=0.0305 \times 931.48 = 28.4 MeV$

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The energy of an eleforn in $n^{th}$ orbit of hydrogen is ${ -13.6 \over n^2 } eV$ energy required to exite the electron form the first orbit $4^{th}$ orbit is

A.

13.6

B.

3.4 eV

C.

0.85eV

D.

1.5 eV

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Explanation

Thomson model

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The activity of a radioactive sample is measured as no counts Per minute at t=o and No counts Per minute t=5 min The time (in min) at which activity reduces to half its value is

A. $ log \,e {2 \over 5} $ B. $ 5 log _{10} 2 $ C. $ 5 log _{e} 2$ D. $ log _{10}^ {2\over5}$

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Explanation

$\triangle E= E_4 - E_1 = {-13.6 \over 4} - ({-13.6 \over 1})= 0.85 +13.6 = 12.75 ev$

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A heavy nucleus at lest breaks into two fragments which fly off with velocities in the ratio 8:1 The ratio of radil of the fragments is

A. 1:02 B. 4:1 C. 1:4 D. 2:1

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Explanation

In mean life (T) Rest mass = $No/ e$ $ T = { 1 \over \lambda } \Rightarrow \lambda = {1 \over 5 } min ^{-1}$ $ N = N0e^{- \lambda t}$ $ N0/2 = N0e^{-1/5 t}$ $m2 = { 1\over 5}t $ $ {1 \over 2 } = e^{-1/5} t $ $ t = 5ln2$ $ 2^{-1} = e^{-1/5} t $ $ t = 5 log e^2$ $ 2^1 = e {1 \over 5 } t $

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The Probability of survival of a radioactive nucleus for one mean life time is

A. $ 1 -{1 \over e} $ B. $ 1 \over e $ C. $ 2 \over e $ D. $ 3 \over e $

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In the nuclear decay below $ _ZX^A \rightarrow _{Z+1}Y^A \rightarrow _{Z-1}B^{A-4} \rightarrow _{Z-1}B^{A-4}$

A. $ \beta,\alpha,\gamma$ B. $ \alpha, \beta, \gamma$ C. $ \gamma, \beta, \alpha $ D. $ \beta,\gamma,\alpha$

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Explanation

$ _ZX^A \rightarrow _{Z+1}Y^A \rightarrow _{Z-1}B^{A-4} \rightarrow _{Z-1}B^{A-4}$

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The wave lenght of the first line of Lyman series for hydrogen atom is equal to that of hydrogen atom is equal to that of second line of Balmar series for a hydrogen like ion. The atomic number Z of hydrogen like ion is

A. 1 B. 2 C. 3 D. 4

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Explanation

$ {1 \over \lambda } = R2^2[{1\over n_2^2} -{1\over n_1^2}] \leftarrow for \, any \,atom $

$ { 1 \over \lambda_1 (L) } = R(1)^2[{1\over 1^2} - {1 \over 2^2}]=RZ^2[{1\over 4} - { 1 \over 16 } ] {3RZ^2 \over 4}$

$ { 1 \over \lambda_1 (L) } = {3R \over 4} \Rightarrow \lambda_1(L) = { 4 \over 3R} $

$ { 1 \over \lambda_2 (B) } = R2^2[{1\over 2^2} - {1 \over 4^2}]=RZ^2[{1\over 4} - { 1 \over 16 } ] = {3RZ^2 \over 16}$

$ \Rightarrow \lambda _2 (B) = { 16 \over 3RZ^2} $ $ { \lambda_1 (L) \over \lambda _2 (B) }$ $ 1 = { 2^2 \over 4} \Rightarrow z^2= 4 Z=2$

Physics MCQs for NEET — Practice Questions with Answers

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