NEET Atomic Structure MCQs & PYQs

The energy of the nth orbit is $E_n = -\frac{13.6}{n^2} eV$. For $n=3$, $E_3 = -1.51 eV$, and for $n=4$, $E_4 = -0.85 eV$. The energy difference is $\Delta E = E_4 - E_3 = -0.85 - (-1.51) = 0.66 eV$.

According to Bohr's second postulate, the electron revolves only in those orbits for which its angular momentum is an integral multiple of $h/(2\pi)$. Hence, it assumes that the angular momentum of electrons is quantized.

The second excited state corresponds to $n=3$. The energy of this state is $E_3 = -\frac{13.6}{3^2} = -1.51 eV$. The energy required to remove the electron to infinity (ionization) is $0 - (-1.51) = 1.51 eV$.

The radius of the nth orbit in a hydrogen-like species is given by $r_n \propto \frac{n^2}{Z}$. For the first orbit ($n=1$), the radius is inversely proportional to the atomic number $Z$. Doubly ionised lithium ($Li^{2+}$) has the maximum $Z=3$, thus it has the minimum radius.

Energy of an electron in a hydrogen-like atom is $E \propto Z^2$. For hydrogen $Z=1$, $E = E_n$. For singly ionised helium ($He^+$), $Z=2$. Therefore, the energy in the same orbit is $2^2 E_n = 4 E_n$.

Frequency $\nu \propto \Delta E$. Emission occurs when transitioning from a higher to a lower energy state. The energy difference between $n=2$ and $n=1$ ($10.2 eV$) is the largest among the given emission transitions.

Transitions ending at $n=2$ belong to the Balmer series. The transition $n=3 \rightarrow 2$ is the first line, and $n=4 \rightarrow 2$ is the second line of the Balmer series.

The necessary centripetal force is provided by the electrostatic force of attraction. Thus, $\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}$. Solving for velocity yields $v = \frac{e}{\sqrt{4\pi\epsilon_0 m r}}$.

The energy of the nth state is given by $E_n = \frac{-13.6}{n^2} eV$. The first excited state corresponds to $n=2$. Therefore, $E_2 = \frac{-13.6}{2^2} = -3.4 eV$.

The radius of the nth Bohr orbit is $r_n = r_0 n^2$, where $r_0$ is the Bohr radius. For the first excited level ($n=2$), $r_2 = r_0(2^2) = 4r_0$. Hence, it is 4 times the Bohr radius.