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Atoms NEET Questions — Free MCQ Practice

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In each of the following question match column -I and column -II select correct Answer. (a) Bohr atom model (b) Ionisation potential (c) Rutherford atom modal (d) Thomson atom modal charge are distrited uniformely. (p) fixed for the atom (q) Nucleus (r) stationary orbits (s) In atom positive and negative

A.

a-->s , b-->r , c -->q , d-->p

B.

a-->r , b-->p , c -->q , d-->s

C.

a -->p , b-->r , c -->s , d-->q

D.

b-->p , c -->q , b-->r , d-->s

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Explanation

Bohr atom model-->stationary Orbits Ionization Potential--> Fixed for the atom Rutherford atom model--> Nucleus Thomson atom model--> In atom positive and Negative charge are distributed uniformly.

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The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy?

A.

|

B.

||

C.

|||

D.

IV

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Explanation

In the given diagram representing energy levels, the transition ||| (III) represents the emission of a photon with the most energy. This is because the energy difference between the levels in transition III is the greatest, which corresponds to the highest energy photon emitted.

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Each of the following question contain two statements.choose correct answer form the given below. statement-I:- Large angle scattering of alpha Particle led to discovery of atomic nucleus. statement-II :- Entire Positive charge of atom is concentrated in the central core.

A. statement -I and II are true. and statement II is correct explanation of statement-I B. statement -I and II are true, but statement-II is not correct explanation of statement I C. statement I is true, but statement II is false. D. statement I is false but statement II is true

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Explanation

Statement I is true because the large angle scattering of alpha particles in Rutherford's gold foil experiment led to the discovery of the atomic nucleus. Statement II is also true because the positive charge of an atom is concentrated in the nucleus. Statement II correctly explains why the large-angle scattering occurs, as the positively charged alpha particles are deflected by the concentrated positive charge in the nucleus.

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Energy leVels A , B , C of a certain atom corresponding values of energy i.e$E_A$<$E_B$<$E_C$ . If $\lambda_1$,$\lambda_2$,$\lambda_3$ are wave lengths of radition corresponding to the transition C-->B , B-->A and C-->A. which of the following is correct

A. $\lambda_3^2 = \lambda_3^2 +\lambda_3^2$ B. $\lambda_3 = \lambda_1 + \lambda_2$ C. $\lambda_3={ \lambda_1\lambda_2 \over\lambda_1+\lambda_2}$ D. $\lambda_1+\lambda_2+\lambda_3= 0$

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Explanation

The minimum energy $\therefore$ $Orbit\; energy E_n=-{13.6\over n^2}= -{13.6\over (1)^2}=-13.6 ev$

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According to Bohr's theory the radius of electron in an orbit described by Principal quantum number n and atomic number Z, is Proportional to.

A. $Z^2n^2$ B. $ { n^2\over z } $ C. ${ Z^2\over n }$ D. $ {n^2\over Z^2 } $

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Explanation

According to Bohr's theory, the radius of an electron in an orbit is given by $r_n = rac{n^2h^2}{4\pi^2m_ekZe^2}$, which simplifies to $r_n \\propto \\frac{n^2}{Z}$. Therefore, the radius is directly proportional to $\frac{n^2}{Z}$. This makes option o2 the correct choice.

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The energy of electron in the nth orbit of hydogen atom is expressed as $E_n ={ - 13.6\over n^2} eV$ . The shortest and longest wave length of lyman series will be.

A. $ 910A ^\circ , 1213A ^\circ $ B. $ 1315 A ^\circ ,1530 A ^\circ $ C. $ 5463 A ^ \circ , 7858 A ^ \circ $ D. None of these

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Explanation

Bohr radius r = $n^2 h^2 \epsilon o \over{ \pi Z e^2 m}$ $\therefore r \alpha{ n^2 \over Z} $

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Number of spectral lines in hydrogen atom is.

A. 6 B. 8 C. 15 D. Infinity

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Explanation

For, shortest wave length in Lyman series $1\over {\lambda min}$ = R[$1\over {1^2}$-$1\over {\alpha ^2}$] => $\lambda$ min = $ 911 A ^\circ $ For, longest wave length in Lyman series $1\over {\lambda max}$ = R[$1\over {1^2}$-$1\over {2^2}$] => $\lambda$ max = $ 1215 A ^ \circ $

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The distance of the closest approach of an alpha particle fired at a nucleus with kinetic energy K is ro. The distance of the closest approach when the $\alpha$- particle is fired at the same nucleus with kinetic energy 2k will be.

A. $r_o \over 2 $ B. $4r_0$ C. $r_o \over 4 $ D. $2r_0$

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Explanation

At a distance of closest approach kinetic energy of $\alpha$ -particle = Potential energy of the system $ {1\over 2} {mv}^2 = K_1 ={Kq_1q_2 \over ro} $ $ \therefore ro = {Kq_1q_2 \over ro} $ When kinetic energy 219 than $ro^1 = {ro \over 2} $

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which of the following series in the spectrum of hydrogen alon lies in the visible legion of the electro magnetic spectrum?

A. Paschen B. Lyman C. Brakett D. Balmer

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Explanation

The Balmer series of hydrogen emission lines lies in the visible region of the electromagnetic spectrum. These lines occur when an electron transitions from a higher energy level to the second energy level (n = 2).

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If 13.6 eV energy is required to ionige the hydrogen atom the energy required to remove the electron form n=2 state is

A. Zero B. 10.2 eV C. 6.8 eV D. 3.4 eV

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Explanation

Inigation potentier (For hydrogen) $E_n = {13.6 \over n^2} eV$

Atoms MCQs for NEET — Physics Questions with Answers

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