NEET Practice Questions (MCQs) with Answers & Solutions

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.

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.

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.

Statement I is false because nuclei of different atoms do not have the same size; their size depends on the mass number (A). Statement II is true because the radius of a nucleus is given by the formula $R = R_0(A)^{1/3}$, where $R_0$ is a constant and $A$ is the mass number. Hence, statement II is correct but does not support statement I.

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 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.

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

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 $

$X (\eta ,\alpha)_3^7 Li $ $_Z^A{X}+_0{n}^1 -->_3^7{Li}+_2{He}^4$ By equating Z and A we get $_Z^A{X} -->_5^{10} {B}$

According to conservation of monentum $m_1v_1 + m_2 v_2 = 0$ $4v_1 + 206 v_2 = 0 $ $ 206 v_2 = - 4 v_2 $ $v_2 = {-4v_1 \over 206}$ Recoil speed of daughter nucleus is $4v \over 206$