Mix Example - Atoms Questions in English

(B) Statement $I$ is correct. Atoms are electrically neutral because the number of protons (positive charge) in the nucleus is equal to the number of electrons (negative charge) orbiting the nucleus.
Statement $II$ is incorrect. While many atoms are stable,there are many radioactive elements whose atoms are unstable and undergo decay. Furthermore,while atoms do emit characteristic spectra,the statement implies that 'each' element's atom is stable,which is a generalization that ignores radioactive instability.

(A) For a hydrogen-like atom,the radius of the $n$-th orbit is $r_n = a_0 \frac{n^2}{Z}$.
The relative change in radius for two consecutive orbitals is $\frac{\Delta r}{r} \approx \frac{dr}{r} = \frac{2n \, dn}{n^2} = \frac{2 \Delta n}{n}$. Since $\Delta n = 1$,$\frac{\Delta r}{r} \propto \frac{1}{n}$. This is independent of $Z$. Thus,$(A)$ and $(B)$ are correct.
The energy of the $n$-th orbit is $E_n = -E_0 \frac{Z^2}{n^2}$.
The relative change in energy is $\frac{\Delta E}{E} \approx \frac{dE}{E} = \frac{2n^{-3} \, dn}{n^{-2}} = \frac{2 \Delta n}{n}$. Since $\Delta n = 1$,$\frac{\Delta E}{E} \propto \frac{1}{n}$. Thus,$(C)$ is incorrect.
The angular momentum is $L = \frac{nh}{2\pi}$.
The relative change is $\frac{\Delta L}{L} = \frac{\Delta n}{n}$. Since $\Delta n = 1$,$\frac{\Delta L}{L} \propto \frac{1}{n}$. Thus,$(D)$ is correct.
Therefore,the correct statements are $(A)$,$(B)$,and $(D)$.

(A) The wavelength $\lambda$ of the emitted photon is given by the Rydberg formula:
$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
Given $n_1 = 1$ and $n_2 = 5$:
$\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{5^2} \right) = R \left( 1 - \frac{1}{25} \right) = \frac{24}{25} R$ ... $(i)$
The momentum of the emitted photon is $p = \frac{h}{\lambda}$.
Substituting from $(i)$:
$p = h \left( \frac{24}{25} R \right)$
By the law of conservation of linear momentum,the momentum of the atom must be equal and opposite to the momentum of the photon (since the atom was initially stationary):
$p_{\text{atom}} = p_{\text{photon}}$
$mv = \frac{24 Rh}{25}$
$v = \frac{24 Rh}{25 m}$

(D) The line emission spectrum is produced by an excited substance in the atomic state,such as a mercury vapour lamp.
Line absorption spectrum is produced when electromagnetic radiations pass through a medium; therefore,the sunlight spectrum is a line absorption spectrum.
The band spectrum is used to study the molecular structure.
Since all the statements $A$,$B$,and $C$ are correct,the correct option is $D$.

(A) An incandescent electric lamp produces a continuous emission spectrum because the filament emits radiation across a wide range of wavelengths due to its high temperature.
Mercury and sodium vapour lamps produce line emission spectra because the atoms in the gas emit light only at specific discrete wavelengths.
Polyatomic substances,such as $H_{2}$,$CO_{2}$,and $KMnO_{4}$,typically produce band absorption spectra.

(D) The solar spectrum consists of a continuous spectrum of light emitted by the hot interior of the Sun,which is crossed by numerous dark lines known as Fraunhofer lines. These dark lines are produced because the cooler gases in the Sun's outer atmosphere (the photosphere and chromosphere) absorb specific wavelengths of light from the continuous spectrum. Therefore,the solar spectrum is classified as a line absorption spectrum.

(B) The energy required to remove the first electron from a neutral helium atom $(He)$ is given as $24.6 eV$.
After the first electron is removed,the remaining ion is $He^+$,which is a hydrogen-like species with atomic number $Z = 2$.
The energy required to remove the second electron from the ground state of $He^+$ is given by the formula $E = Z^2 \times 13.6 eV$.
Substituting $Z = 2$,we get $E = (2)^2 \times 13.6 eV = 4 \times 13.6 eV = 54.4 eV$.
The total energy required to remove both electrons is the sum of the energy to remove the first and the second electron.
Total Energy $= 24.6 eV + 54.4 eV = 79 eV$.

(A) The energy of the photon emitted in the first line of the Lyman series ($n=2$ to $n=1$) is given by: $E = 13.6 \ eV \times (1 - 1/4) = 13.6 \times 0.75 = 10.2 \ eV$.
Converting this to Joules: $E = 10.2 \times 1.6 \times 10^{-19} \ J = 1.632 \times 10^{-18} \ J$.
The momentum of the photon is $p = E/c = (1.632 \times 10^{-18}) / (3 \times 10^8) = 5.44 \times 10^{-27} \ kg \ m/s$.
By conservation of momentum,the recoil momentum of the hydrogen atom must be equal and opposite to the photon's momentum: $m_H v = p$.
The mass of a hydrogen atom is $m_H \approx 1.67 \times 10^{-27} \ kg$.
Therefore,$v = p / m_H = (5.44 \times 10^{-27}) / (1.67 \times 10^{-27}) \approx 3.25 \ m/s$.

(D) The correct matches are as follows:
$1$. $A$ burning candle emits light across a wide range of wavelengths,resulting in a continuous spectrum.
$2$. Sodium vapour,when excited,emits light at specific discrete wavelengths,resulting in a line spectrum.
$3$. $A$ Bunsen flame (often containing molecular species) produces a band spectrum.
$4$. Dark lines in the solar spectrum (Fraunhofer lines) are caused by the absorption of specific wavelengths by gases in the solar atmosphere,resulting in an absorption spectrum.
Therefore,the correct matching is $1-(ii), 2-(i), 3-(iii), 4-(iv)$.

(A) $1$. Nitrogen molecules exhibit a band spectrum because molecular spectra consist of bands due to transitions between electronic,vibrational,and rotational energy levels.
$2$. Incandescent solids emit a continuous spectrum because they contain a wide range of frequencies due to thermal agitation.
$3$. Fraunhofer lines are dark lines observed in the solar spectrum,which are caused by the absorption of specific wavelengths by gases in the solar atmosphere,thus forming an absorption spectrum.
$4$. An electric arc between iron rods produces a line emission spectrum characteristic of the iron atoms present in the arc.
Therefore,the correct matching is $1-C, 2-A, 3-B, 4-D$.

(D) Given,kinetic energy of the electron $= 5.5 eV$.
The energy required to excite a hydrogen atom from the ground state $(n=1)$ to the first excited state $(n=2)$ is $\Delta E = E_2 - E_1 = -3.4 eV - (-13.6 eV) = 10.2 eV$.
Since the kinetic energy of the incident electron $(5.5 eV)$ is less than the excitation energy required $(10.2 eV)$,the electron cannot transfer any energy to the hydrogen atom to cause an electronic transition.
As no energy is lost to internal excitation,the total kinetic energy of the system remains conserved.
Therefore,the collision is elastic.