NEET Physics Questions: Atoms

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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





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.

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?





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.
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.




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.
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




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




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.




Bohr radius r = $n^2 h^2 \epsilon o \over{ \pi Z e^2 m}$ $\therefore r \alpha{ n^2 \over Z} $
Number of spectral lines in hydrogen atom is.




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 $
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.




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




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




Inigation potentier (For hydrogen) $E_n = {13.6 \over n^2} eV$
If the binding energy of electron in a hydrogen atom is 13.6 eV, the energy required to remove the electron form the first state of $Li^{2+}$ is.




Binding energy = ${13.6 Z^2 \over n^2}$ For $Li^{2+}$ Z = 3, n = 2 first exited state
The ionization Potential of hydrogen atom is 13.6 eV. An electron in the ground state absorbs Photon of energy 12.75 eV. How many different spectral lines can one expect when electron make a down ward transition




$\triangle E = E_n - E_1$ $E_n = \triangle E + E_1 = 12.75-13.6 = -0.85$ $E_n = -{13.6 \over n^2}$ $ -0.85 = {-13.6 \over n^2} $ $ n^2 = {13.6 \over 0.85} = 16 --> n=4 $

An $\alpha$ -particle of energy ${1\over2} mv^2$ bombards by a heavy nuclear target ofcharge ze.Then the distance of closet approach for the alpha nucleus will be Proportional to





at distance of closest approach K.E = P.E ${1\over 2}mv^2 = {1 \over 4 \pi \epsilon_o} {(ze) (ze) \over ro} \Rightarrow ro = {ze^2 \over \pi epsilon_o mv^2 }$

An $\alpha $ -particle of energy 5 MeV is scattered though 180 by a fixed uranium nucleus. The distance of the closest approach nucleus The distance of the closest approach is of the order of




The distance of closest approach in a Rutherford scattering experiment can be estimated using the formula: $$ d = rac{1}{4 \\pi \\epsilon_0} \\frac{2Ze^2}{E} $$ where $Z$ is the atomic number of the uranium nucleus (92), $e$ is the elementary charge, and $E$ is the energy of the alpha particle (5 MeV). Substituting these values, the order of magnitude for the distance of closest approach comes out to be approximately $10^{-12}$ cm.
which of the following transition in hydrogen atoms emits Photon of highest frequency?




n = 2 to n = 6 absorbs photon n = 9 to n = 2 absorbs photon n = 6 to n = 2 emission of photon n = 2 to n = 1 emission of photon $E = E_2-E_1 = {-13.6 \over 36}-({-13.6 \over 4 }) = -0.38+13.6 = 3.02 eV $ $ \triangle \Sigma^1 = E_2-E_1 = {-13.6 \over 4}-({-13.6 \over 1}) = -3.4+13.6 = 10.2 eV $ $ \triangle \Sigma^1 > \triangle E$ $hf' > hf $