Atomic models and Planck's quantum theory Questions in English

(A) For an electron in a Bohr orbit,the potential energy $(P.E.)$ is given by $P.E. = -\frac{kZe^2}{r}$.
The kinetic energy $(K.E.)$ is given by $K.E. = \frac{kZe^2}{2r}$.
Comparing the two expressions,we find that $K.E. = -\frac{1}{2} P.E.$
Therefore,the magnitude of kinetic energy is equal to half of the magnitude of potential energy,i.e.,$|K.E.| = \frac{1}{2} |P.E.|$.

(C) The most probable radius for an electron in a hydrogen-like species is given by the formula $r_{mp} = \frac{a_0}{Z}$, where $a_0$ is the Bohr radius $(52.9 \ pm)$ and $Z$ is the atomic number.
For the helium ion $(He^{+})$, the atomic number $Z = 2$.
Substituting the values: $r_{mp} = \frac{52.9 \ pm}{2} = 26.45 \ pm$.
Rounding to the nearest value, we get $26.5 \ pm$.

(B) Rutherford's $\alpha$-particle scattering experiment demonstrated that most of the atom is empty space,with a small,dense,positively charged nucleus at the center. This led to the conclusion that electrons revolve around the nucleus in the extra-nuclear region. Therefore,the correct option is $(B)$.

(B) Bohr's model is applicable only to hydrogen-like species,which are atoms or ions containing only one electron,such as $H$,$He^+$,$Li^{2+}$,etc.

(D) . The nucleus occupies a much smaller volume compared to the total volume of the atom.
Since the nucleus is extremely small and positively charged,most of the space in an atom is empty,allowing the majority of alpha particles to pass through without any deflection.

(B) The Rydberg constant $R$ is proportional to the reduced mass $\mu$ of the system.
For a hydrogen-like atom,the reduced mass is $\mu = \frac{m_e M}{m_e + M}$,where $M$ is the mass of the nucleus. Since $M \gg m_e$,$\mu \approx m_e$,and $R_\infty = \frac{m_e e^4}{8 \epsilon_0^2 h^3 c}$.
For positronium,the system consists of an electron $(m_e)$ and a positron $(m_p = m_e)$.
The reduced mass $\mu$ for positronium is:
$\mu = \frac{m_e \times m_p}{m_e + m_p} = \frac{m_e \times m_e}{m_e + m_e} = \frac{m_e^2}{2m_e} = \frac{m_e}{2}$.
Since $R \propto \mu$,the Rydberg constant for positronium $(R_{pos})$ is related to the Rydberg constant for an infinite nuclear mass $(R_\infty)$ as:
$R_{pos} = \frac{\mu}{m_e} R_\infty = \frac{m_e / 2}{m_e} R_\infty = \frac{R_\infty}{2}$.

(C) The correct answer is $(C)$.
In the Rutherford $\alpha$-particle scattering experiment,it was observed that most of the $\alpha$-particles passed through the gold foil without any deflection.
This indicates that the majority of the space within an atom is empty.

(B) The $K$ shell is the first shell $(n=1)$ and the $L$ shell is the second shell $(n=2)$.
Since the $K$ shell is closer to the nucleus,it has lower energy than the $L$ shell.
When an electron transitions from a higher energy level ($L$ shell) to a lower energy level ($K$ shell),the difference in energy is emitted as a photon.
Therefore,energy is released.

(A) According to Planck's quantum theory,energy is always absorbed or emitted in discrete amounts,which are whole number multiples of a quantum $(E = nh\nu)$.
Therefore,the statement that energy is not absorbed or emitted in whole number or multiple of quantum is incorrect.

(A) The energy of an electron in an orbit is given by the formula $E_n = -13.6 \times \frac{Z^2}{n^2} \text{ eV}$.
As we move away from the nucleus,the principal quantum number $n$ increases.
Since $E_n$ is inversely proportional to $n^2$ and has a negative sign,as $n$ increases,the value of $E_n$ becomes less negative,which means the energy increases.

(D) The $Bohr$ model was successful in explaining the line spectrum of hydrogen,including emission and absorption spectra. However,it failed to explain the fine structure of spectral lines observed in high-resolution spectroscopy,which arises due to the splitting of energy levels.

(B) The existence of a positively charged nucleus was established by Ernest Rutherford through his famous $\alpha$-particle scattering experiment (also known as the Geiger-Marsden experiment). In this experiment,he bombarded a thin gold foil with $\alpha$-particles and observed that a small fraction of particles were deflected at large angles,leading to the conclusion that the positive charge and mass of the atom are concentrated in a very small central region called the nucleus.

(C) The wavelength of a spectral line for an electronic transition is inversely proportional to the difference in the energy involved in the transition.
According to the relation $\Delta E = E_2 - E_1 = \frac{hc}{\lambda}$,where $\Delta E$ is the energy difference,$h$ is Planck's constant,$c$ is the speed of light,and $\lambda$ is the wavelength.
Thus,$\lambda = \frac{hc}{\Delta E}$,which implies $\lambda \propto \frac{1}{\Delta E}$.

(A) When an electron transitions from a higher energy level $(n_2)$ to a lower energy level $(n_1)$,the difference in energy is released in the form of a photon.
The energy of the emitted photon is given by the equation $E = \Delta E = E_2 - E_1 = hc / \lambda$.
Conversely,when an electron moves from a lower energy level to a higher energy level,it must absorb energy equal to the difference between the two levels.
The state with the lowest energy for an electron is called the ground state.

(A) When an electron moves from a lower energy level $(n_1)$ to a higher energy level $(n_2)$,it must absorb a photon of energy equal to the difference between the two levels,$\Delta E = E_2 - E_1$.
Therefore,the energy of the electron increases.

(C) Fast-moving electrons are called $\beta$-particles.
They pass through the tin metal foil because most of the atom is empty space,as explained by Rutherford in his gold foil experiment.

(D) The energy of an orbit in a hydrogen atom is given by the formula $E_n = \frac{-1312}{n^2} \ kJ \ mol^{-1}$.
Given,for the second orbit $(n=2)$,$E_2 = -328 \ kJ \ mol^{-1}$.
For the fourth orbit $(n=4)$,the energy $E_4$ is related to $E_2$ by the ratio $\frac{E_4}{E_2} = \frac{n_2^2}{n_4^2} = \frac{2^2}{4^2} = \frac{4}{16} = \frac{1}{4}$.
Therefore,$E_4 = \frac{E_2}{4} = \frac{-328}{4} = -82 \ kJ \ mol^{-1}$.

(D) According to Bohr's atomic model,electrons revolve in specific orbits called stationary orbits. While revolving in these orbits,the electron does not lose or gain energy. Therefore,its energy remains constant.

(A) According to Bohr's theory,electrons revolve in stationary orbits with fixed energy levels. When an electron moves from one orbit to another,it does so by absorbing or emitting a photon of energy equal to the difference between the two energy levels. This process is known as a quantum jump or transition,which implies that the electron does not exist in the space between the orbits. This concept is fundamentally based on the $quantisation$ of energy,where energy levels are discrete and not continuous.

(B) According to the quantum theory of radiation,a hot body emits radiant energy not continuously but discontinuously in the form of small packets of energy called quanta or photons.

(C) According to Bohr's model for the hydrogen atom,the energy of an electron in the $n^{th}$ orbit is given by the formula:
$E_n = -\frac{13.6}{n^2} \ eV$
where $n$ is the principal quantum number representing the orbit number.

(A) According to Bohr's theory,the radius of the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $r_n = \frac{n^2 h^2}{4 \pi^2 m k Z e^2}$.
Assuming the constant $k = \frac{1}{4 \pi \epsilon_0}$,the expression simplifies to $r = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2}$.
However,in the given options,the standard representation for the radius of the $n^{th}$ orbit is $r = \frac{n^2 h^2}{4 \pi^2 m Z e^2}$ (where $k$ is implicitly included or units are in $CGS$). Thus,option $A$ and $B$ represent the correct physical relationship.

(C) According to Bohr's theory,the total energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula:
$E_n = - \frac{2\pi^2 m e^4 z^2}{n^2 h^2}$
where $m$ is the mass of the electron,$e$ is the charge of the electron,$z$ is the atomic number,$n$ is the principal quantum number,and $h$ is Planck's constant.
Comparing this with the given options,option $C$ represents the correct expression.

(D) Arnold $Sommerfeld$ modified $Bohr's$ atomic model in $1916$. He proposed that electrons move in elliptical orbits around the nucleus,in addition to the circular orbits proposed by $Bohr$. This modification helped explain the fine structure of spectral lines in hydrogen-like atoms.

(B) The expression for Bohr's radius is given by $r_n = \frac{n^2 h^2}{4 \pi^2 m e^2 Z}$.
Since $n$,$h$,$\pi$,$m$,$e$,and $Z$ are all constants or positive integers,the radius $r_n$ must always be a positive ($+$ve) value.

(B) It was in $1913$ that Neils Bohr put forth the stability of the atom and with the help of Planck's quantum theory explained the reason for spectral lines.
Bohr first made use of quantum theory to explain the structure of atoms and proposed that the energy of electrons in an atom is quantized.

(C) The energy of an electron in an orbit is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$. As the electron moves away from the nucleus,the principal quantum number $n$ increases. Since the energy expression is negative,as $n$ increases,the value of $E_n$ becomes less negative,which means the total energy of the electron increases.

(C) The conclusions of Rutherford from his $\alpha$-particle scattering experiment are:
$(i)$ Most of the space inside the atom is empty because most of the $\alpha$-particles passed through the gold foil without getting deflected.
$(ii)$ Very few particles were deflected from their path,indicating that the positive charge of the atom occupies a very small volume,known as the nucleus.
$(iii)$ $A$ very small fraction of $\alpha$-particles were deflected by $180^{\circ}$,which confirmed that the positive charge is concentrated in a very small space.

(A) In his famous $\alpha$-particle scattering experiment,Ernest Rutherford used a thin foil of $Gold$ $(Au)$ to study the structure of the atom.
Therefore,the correct option is $(A)$.

(C) The energy emitted when an electron transitions between energy levels in a hydrogen atom is given by the formula: $\Delta E = 13.6 \, \text{eV} \times \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.
Given $n_i = 3$ and $n_f = 2$,we have:
$\Delta E = 13.6 \times \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \, \text{eV}$.
$\Delta E = 13.6 \times \left( \frac{1}{4} - \frac{1}{9} \right) \, \text{eV}$.
$\Delta E = 13.6 \times \left( \frac{9 - 4}{36} \right) \, \text{eV}$.
$\Delta E = 13.6 \times \frac{5}{36} \, \text{eV} \approx 1.89 \, \text{eV}$,which rounds to $1.9 \, \text{eV}$.

(D) The main postulate of Bohr's model is that the electron can move only in those circular orbits where its angular momentum is an integral multiple of $\frac{h}{2\pi}$.
This is expressed as $mvr = n \frac{h}{2\pi}$,where $n = 1, 2, 3, ...$ (an integer).

(B) In Bohr's theory,the circular paths in which electrons revolve around the nucleus are known as $Orbits$ or $Stationary$ $states$.

(C) The spectral lines of the hydrogen atom are formed by electronic transitions between energy levels.
The transitions from higher energy levels $(n_2 > 2)$ to the second energy level $(n_1 = 2)$ result in the emission of radiation with wavelengths ranging from approximately $365 \, nm$ to $656 \, nm$.
These wavelengths fall within the visible region of the electromagnetic spectrum and constitute the $Balmer$ series.

(D) The radius of the $n^{th}$ Bohr orbit is given by the formula: $r_n = \frac{n^2 \times 0.529 \mathring{A}}{Z}$.
For the hydrogen atom,the atomic number $Z = 1$ and for the first orbit,$n = 1$.
Substituting these values: $r_1 = \frac{1^2 \times 0.529 \mathring{A}}{1} = 0.529 \mathring{A}$.
Rounding to two decimal places,we get $0.53 \mathring{A}$.
Therefore,the correct option is $(d)$.

(A) In the Balmer series,the transitions occur to the $n_1 = 2$ energy level.
The first line corresponds to $n_2 = 3 \rightarrow n_1 = 2$.
The second line corresponds to $n_2 = 4 \rightarrow n_1 = 2$.
The third line corresponds to $n_2 = 5 \rightarrow n_1 = 2$.

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $E_n = - \frac{2.178 \times 10^{-18} \ J}{n^2}$.
For the second Bohr orbit,$n = 2$.
Substituting the value of $n$ into the formula:
$E_2 = - \frac{2.178 \times 10^{-18}}{2^2} \ J$
$E_2 = - \frac{2.178 \times 10^{-18}}{4} \ J$
$E_2 = - 5.445 \times 10^{-19} \ J$.
Thus,the energy is approximately $- 5.44 \times 10^{-19} \ J$.

(C) The relationship between energy and wavelength is given by the formula: $\Delta E = \frac{hc}{\lambda}$.
Rearranging for wavelength: $\lambda = \frac{hc}{\Delta E}$.
Substituting the given values: $\lambda = \frac{6.64 \times 10^{-34} \ J \cdot s \times 3 \times 10^8 \ m/s}{3 \times 10^{-8} \ J}$.
$\lambda = 6.64 \times 10^{-18} \ m$.
Since $1 \ m = 10^{10} \ \mathring{A}$,we have $\lambda = 6.64 \times 10^{-18} \times 10^{10} \ \mathring{A} = 6.64 \times 10^{-8} \ \mathring{A}$.

(D) The radius of the $n^{th}$ Bohr orbit is given by the formula $r_n = r_1 \times n^2$,where $r_1$ is the radius of the first orbit and $n$ is the principal quantum number.
For the third Bohr orbit,$n = 3$.
Substituting the values: $r_3 = 0.53 \ \mathring{A} \times (3)^2$.
$r_3 = 0.53 \times 9 = 4.77 \ \mathring{A}$.

(B) The energy of a photon is given by the equation $E = \frac{hc}{\lambda}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda$ is the wavelength.
From this relation,it is clear that $\lambda = \frac{hc}{E}$.
Therefore,the wavelength of the emitted spectral line is inversely proportional to the energy of the photon,i.e.,$\lambda \propto \frac{1}{E}$.

(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $E_n = \frac{-13.6}{n^2} \ eV$.
For the ground state,$n = 1$.
For the first excited state,$n = 2$.
Substituting $n = 2$ into the formula:
$E_2 = \frac{-13.6}{2^2} = \frac{-13.6}{4} = -3.40 \ eV$.

(B) In the emission spectra of the hydrogen atom,the spectral series are defined by the value of $n_1$ (the lower energy level to which the electron transitions).
For the Lyman series,$n_1 = 1$.
For the Balmer series,$n_1 = 2$.
For the Paschen series,$n_1 = 3$.
Therefore,for the Paschen series,the transition occurs from higher energy states $(n_2 = 4, 5, 6, \dots)$ to the third energy state $(n_1 = 3)$.

(C) According to Bohr's model for a hydrogen atom,the kinetic energy $(K.E.)$ and total energy $(T.E.)$ are related as $K.E. = -T.E.$
Therefore,the ratio of kinetic energy to total energy is $K.E. / T.E. = 1 / -1$.
Thus,the ratio is $1:-1$.

(C) According to Bohr's theory,the energy of an electron in the $n^{th}$ orbit of a Hydrogen atom is given by the formula:
$E_n = - \frac{R_H}{n^2} \ J \ atom^{-1}$
where $R_H$ is the Rydberg constant for energy,which is approximately $2.18 \times 10^{-18} \ J$.
Converting this to $kJ \ mol^{-1}$:
$E_n = - \frac{2.18 \times 10^{-18} \times 6.022 \times 10^{23}}{n^2} \ J \ mol^{-1} \approx - \frac{1312 \ kJ \ mol^{-1}}{n^2}$.
Among the given options,the value closest to the standard constant is $1313.3 \ kJ \ mol^{-1}$.

(B) The radius of the $n^{th}$ Bohr orbit is given by the formula $r_n = a_0 \times n^2$,where $a_0$ is the Bohr radius and $n$ is the principal quantum number.
For the first orbit $(n_1 = 1)$,$r_1 = a_0 \times (1)^2 = a_0$.
For the second orbit $(n_2 = 2)$,$r_2 = a_0 \times (2)^2 = 4a_0$.
The ratio of the radius of the second orbit to the first orbit is $\frac{r_2}{r_1} = \frac{4a_0}{a_0} = 4$.

(B) The wavenumber $\bar{\nu}$ is given by the Rydberg formula: $\bar{\nu} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the transition $n_2 = 2$ to $n_1 = 1$,$\bar{\nu} = 109677 \ cm^{-1} \times (1 - 1/4) = 109677 \times 0.75 = 82257.75 \ cm^{-1}$.
Converting to meters: $\bar{\nu} = 8225775 \ m^{-1}$.
The frequency $\nu$ is calculated as $\nu = c \times \bar{\nu}$.
$\nu = (3 \times 10^8 \ m/s) \times (8225775 \ m^{-1}) = 2.4677 \times 10^{15} \ Hz \approx 24.66 \times 10^{14} \ Hz$.

(B) The radius of an electron in a hydrogen atom is given by the formula $r_n = n^2 a_0$,where $n$ is the principal quantum number and $a_0$ is the Bohr radius.
For the ground state,$n = 1$.
The first excited state corresponds to $n = 2$.
Substituting $n = 2$ into the formula: $r_2 = (2)^2 a_0 = 4a_0$.
Therefore,the correct option is $(B)$.

(D) The radius of the $n^{th}$ orbit is given by $r_n \propto n^2$.
The area of the orbit is $A = \pi r_n^2$.
Substituting $r_n$,we get $A_n \propto (n^2)^2 = n^4$.
Therefore,the ratio of the area of the second orbit $(n=2)$ to the first orbit $(n=1)$ is:
$\frac{A_2}{A_1} = \frac{n_2^4}{n_1^4} = \frac{2^4}{1^4} = \frac{16}{1} = 16:1$.

(A) The time period $T$ for one revolution is given by $T = \frac{2\pi r}{v}$.
From Bohr's postulate of angular momentum,$mvr = \frac{nh}{2\pi}$,which gives $v = \frac{nh}{2\pi mr}$.
Substituting the value of $v$ in the expression for $T$:
$T = \frac{2\pi r}{(nh / 2\pi mr)} = \frac{2\pi r \times 2\pi mr}{nh} = \frac{4\pi^2 mr^2}{nh}$.

(D) The radius of an orbit in a hydrogen-like species is given by the formula: $r_n = 0.529 \times \frac{n^2}{Z} \ \mathring{A}$.
For the first Bohr orbit of the hydrogen atom $(H)$: $n = 1$ and $Z = 1$. Thus,$r_H = 0.529 \times \frac{1^2}{1} = 0.529 \ \mathring{A}$.
Now,checking the options:
For $Be^{3+}$: $Z = 4$ and $n = 2$. Thus,$r = 0.529 \times \frac{2^2}{4} = 0.529 \times \frac{4}{4} = 0.529 \ \mathring{A}$.
Since the radius of $Be^{3+}$ $(n = 2)$ is equal to the radius of the first Bohr orbit of hydrogen,the correct option is $D$.

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{R_H}{n^2}$,where $R_H = 2.18 \times 10^{-18} \ J$.
The energy difference for the transition from $n_2 = 4$ to $n_1 = 1$ is $\Delta E = E_4 - E_1 = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
$\Delta E = 2.18 \times 10^{-18} \left( \frac{1}{1^2} - \frac{1}{4^2} \right) = 2.18 \times 10^{-18} \left( 1 - \frac{1}{16} \right) = 2.18 \times 10^{-18} \times \frac{15}{16} \ J$.
Since $\Delta E = h\nu$,the frequency $\nu = \frac{\Delta E}{h}$.
$\nu = \frac{2.18 \times 10^{-18} \times 15}{16 \times 6.625 \times 10^{-34}} = \frac{32.7 \times 10^{-18}}{106 \times 10^{-34}} \approx 3.08 \times 10^{15} \ s^{-1}$.