AP EAMCET 2021 Chemistry Question Paper with Answer and Solution

(B) The balanced chemical equation is: $CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$
Molar mass of $CaCO_3 = 100 \ g/mol$
Molar mass of $HCl = 36.5 \ g/mol$
Molar mass of $CO_2 = 44 \ g/mol$
Moles of $CaCO_3 = \frac{20}{100} = 0.2 \ mol$
Moles of $HCl = \frac{20}{36.5} = 0.548 \ mol$
According to the reaction,$1 \ mol$ of $CaCO_3$ requires $2 \ mol$ of $HCl$.
So,$0.2 \ mol$ of $CaCO_3$ requires $0.2 \times 2 = 0.4 \ mol$ of $HCl$.
Since we have $0.548 \ mol$ of $HCl$,$CaCO_3$ is the limiting reagent.
Mass of $CO_2$ formed = $0.2 \ mol \times 44 \ g/mol = 8.8 \ g$.

(C) Step $1$: Calculate the mass of one mole of element $X$ (molar mass).
Mass of one atom = $6.643 \times 10^{-23} \ g$.
Molar mass = Mass of one atom $\times$ Avogadro's number $(N_A)$.
Molar mass = $(6.643 \times 10^{-23} \ g) \times (6.022 \times 10^{23} \ mol^{-1}) \approx 40 \ g/mol$.
Step $2$: Calculate the number of moles in $50 \ kg$ $(50,000 \ g)$.
Number of moles = $\frac{\text{Given mass}}{\text{Molar mass}} = \frac{50,000 \ g}{40 \ g/mol} = 1250 \ moles$.

(D) $1$. Calculate the mass of pure $H_2SO_4$ required:
$n = M \times V(L) = 0.5 \ mol \ L^{-1} \times 0.5 \ L = 0.25 \ mol$.
Mass of $H_2SO_4 = n \times \text{molar mass} = 0.25 \ mol \times 98.079 \ g \ mol^{-1} = 24.51975 \ g \approx 24.52 \ g$.
$2$. Calculate the mass of the $90\%$ solution required:
Since the solution is $90\%$ pure,$\text{Mass of solution} = \frac{\text{Mass of solute}}{\text{Percentage}} = \frac{24.51975 \ g}{0.90} = 27.244 \ g \approx 27.24 \ g$.

(C) The balanced chemical equation for the formation of ammonia is:
$N_2(g) + 3H_2(g) \longrightarrow 2NH_3(g)$
According to the stoichiometry:
$1 \ mol$ of $N_2$ ($22.4 \ L$ at $NTP$) reacts with $3 \ mol$ of $H_2$ ($3 \times 22.4 \ L = 67.2 \ L$ at $NTP$) to produce $2 \ mol$ of $NH_3$ ($2 \times 22.4 \ L = 44.8 \ L$ at $NTP$).
The molar mass of $NH_3$ is $17 \ g/mol$. Therefore,$2 \ mol$ of $NH_3$ corresponds to $2 \times 17 = 34 \ g$.
For $34 \ g$ of $NH_3$,we require $67.2 \ L$ of $H_2$ and $22.4 \ L$ of $N_2$.
For $3.40 \ g$ of $NH_3$ (which is $0.1$ times $34 \ g$),the required volumes are:
Volume of $H_2 = 0.1 \times 67.2 \ L = 6.72 \ L$
Volume of $N_2 = 0.1 \times 22.4 \ L = 2.24 \ L$
Thus,the required volumes are $6.72 \ L$ of $H_2$ and $2.24 \ L$ of $N_2$.

(B) The balanced chemical equation for the formation of ammonia is:
$N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$
From the stoichiometry of the reaction:
$2 \text{ moles}$ of $NH_3$ are produced from $3 \text{ moles}$ of $H_2$.
Molar mass of $NH_3 = 17 \text{ g/mol}$.
Molar mass of $H_2 = 2 \text{ g/mol}$.
Therefore,$2 \times 17 \text{ g} = 34 \text{ g}$ of $NH_3$ is produced from $3 \times 2 \text{ g} = 6 \text{ g}$ of $H_2$.
To produce $100 \text{ g}$ of $NH_3$,the mass of $H_2$ required is:
$\text{Mass of } H_2 = \frac{6 \text{ g}}{34 \text{ g}} \times 100 \text{ g} = 17.647 \text{ g} \approx 17.65 \text{ g}$.

(C) The balanced chemical equation is: $Al(OH)_3 + 3HCl \longrightarrow AlCl_3 + 3H_2O$.
The molar mass of $Al(OH)_3 = 27 + 3 \times (16 + 1) = 78 \ g/mol$.
The molar mass of $HCl = 1 + 35.5 = 36.5 \ g/mol$.
Total mass of $HCl$ produced per day $= 3.0 \ g/L \times 2.6 \ L = 7.8 \ g$.
From the stoichiometry,$3 \times 36.5 = 109.5 \ g$ of $HCl$ is neutralized by $78 \ g$ of $Al(OH)_3$.
Therefore,$7.8 \ g$ of $HCl$ is neutralized by $\frac{78}{109.5} \times 7.8 = 5.556 \ g$ of $Al(OH)_3$.
Each tablet contains $400 \ mg = 0.4 \ g$ of $Al(OH)_3$.
Number of tablets required $= \frac{5.556 \ g}{0.4 \ g/tablet} = 13.89 \approx 14$ tablets.

(B) We know that concentration is dependent on the number of moles of solute and the volume of solution in $L$ and it is given by:
Molarity $= \frac{\text{number of moles of solute}}{\text{volume of solution in } L}$
Here,$n = 0.2 \ mol$,$V = 250 \ mL = 0.25 \ L$
Therefore,
Molarity $= \frac{0.2}{0.25} = 0.8 \ M$
Hence,the concentration of the solution containing $0.2 \ mol$ of sulphuric acid in $250 \ mL$ of water will be $0.8 \ M$.

(A) The dissociation of $CaCl_2$ is given by: $CaCl_2 \rightarrow Ca^{2+} + 2Cl^-$.
One mole of $CaCl_2$ produces $2$ moles of chloride ions $(Cl^-)$.
Number of moles of $Cl^-$ ions $= \frac{3.01 \times 10^{22}}{6.02 \times 10^{23}} = 0.05 \ mol$.
Since $2$ moles of $Cl^-$ come from $1$ mole of $CaCl_2$,the moles of $CaCl_2 = \frac{0.05}{2} = 0.025 \ mol$.
Molarity $(M) = \frac{\text{moles of solute}}{\text{volume of solution in } L} = \frac{0.025 \ mol}{0.5 \ L} = 0.05 \ M$.
Thus,the molarity of the solution is $0.05 \ M$.

(A) Mass of metal $X = 12 \ g \times 0.20 = 2.4 \ g$.
Mass of metal $Y = 12 \ g - 2.4 \ g = 9.6 \ g$.
Moles of $X = \frac{2.4 \ g}{40 \ g/mol} = 0.06 \ mol$.
Since the ratio of atoms $X:Y$ is $2:5$,the ratio of moles $n_X:n_Y$ is also $2:5$.
$n_Y = n_X \times \frac{5}{2} = 0.06 \times 2.5 = 0.15 \ mol$.
Atomic mass of $Y = \frac{\text{Mass of } Y}{n_Y} = \frac{9.6 \ g}{0.15 \ mol} = 64 \ g/mol$.
Thus,the atomic mass of $Y$ is $64 \ amu$.

(C) Molar mass of oxalic acid $(COOH)_2 \cdot 2H_2O = 126 \ g/mol$.
Equivalent mass of oxalic acid = $\frac{\text{Molar mass}}{n\text{-factor}} = \frac{126}{2} = 63 \ g/eq$.
Number of gram equivalents = $\frac{\text{Given mass}}{\text{Equivalent mass}} = \frac{0.63}{63} = 0.01 \ eq$.
Normality $(N) = \frac{\text{Number of gram equivalents}}{\text{Volume of solution in } L} = \frac{0.01}{250/1000} = \frac{0.01 \times 1000}{250} = 0.04 \ N$.
Therefore,the normality of the solution is $0.04 \ N$.

(B) Mass of the carbon tetrachloride $= 15.9 \ g$
Volume of carbon tetrachloride $= 10 \ mL$
Density of carbon tetrachloride $= \frac{\text{Mass}}{\text{Volume}}$
Density of carbon tetrachloride $= \frac{15.9 \ g}{10 \ mL} = 1.59 \ g \cdot mL^{-1}$
Hence,the density of carbon tetrachloride is $1.59 \ g \cdot mL^{-1}$.

(C) For $H_2O_2$,the relationship between Normality $(N)$ and Molarity $(M)$ is given by: $N = n \times M$,where $n$-factor for $H_2O_2$ is $2$.
$M = \frac{N}{2} = \frac{1.5}{2} = 0.75 \ M$.
$\% \left(\frac{W}{V}\right)$ is defined as the mass of solute in grams present in $100 \ mL$ of solution.
Mass of $H_2O_2$ in $1 \ L$ $(1000 \ mL)$ $= M \times \text{Molar Mass} = 0.75 \times 34 = 25.5 \ g/L$.
Therefore,in $100 \ mL$,the mass is $\frac{25.5}{1000} \times 100 = 2.55 \ g$.
Thus,the $\% \left(\frac{W}{V}\right)$ is $2.55$.

(B) The mass percentage of an element in a compound is calculated as: $\text{Mass } \% = \frac{\text{Total mass of the element}}{\text{Molar mass of the compound}} \times 100$.
In $CCl_4$,the molar mass is $12.01 + 4 \times 35.45 = 153.81 \approx 154 \ g/mol$.
Mass percentage of $C = \frac{12.01}{153.81} \times 100 \approx 7.81 \%$.
Mass percentage of $Cl = \frac{4 \times 35.45}{153.81} \times 100 \approx 92.19 \%$.
Given the options provided,the closest values are $7.84$ and $92.80$.

(C) The normality $(N)$ of $H_2O_2$ is related to its molarity $(M)$ by the equation: $N = M \times n$-factor.
For $H_2O_2$,the $n$-factor is $2$,so $M = \frac{N}{2} = \frac{3}{2} = 1.5$ $M$.
The volume strength of $H_2O_2$ is given by the formula: $\text{Volume strength} = M \times 11.2$.
Therefore,$\text{Volume strength} = 1.5 \times 11.2 = 16.8$ $L$.
This is approximately $17$ $L$.

(A) Given,mass of $CO_2 = 0.535 \ g$ and mass of $H_2O = 0.13 \ g$.
Mass of compound $= 0.765 \ g$.
Percentage of $C = \frac{12}{44} \times \frac{0.535}{0.765} \times 100 \approx 19.07 \%$.
Percentage of $H = \frac{2}{18} \times \frac{0.13}{0.765} \times 100 \approx 1.88 \% \approx 2.00 \%$.
Ratio of percentage of $C$ and $H = 19.07 : 1.88 \approx 19 : 2$.

(A) According to the ideal gas equation,we have:
$PV = nRT$
Since the number of moles $n = \frac{m}{M}$,where $m$ is the mass and $M$ is the molar mass,we can substitute this into the equation:
$PV = \frac{m}{M} RT$
Rearranging the terms to solve for density $\rho = \frac{m}{V}$:
$P = \frac{m}{V} \times \frac{RT}{M}$
$P = \rho \times \frac{RT}{M}$
Therefore,the density $\rho$ is given by:
$\rho = \frac{PM}{RT}$

(B) According to Boyle's Law,for a fixed amount of gas at a constant temperature,the pressure $(P)$ is inversely proportional to the volume $(V)$ of the gas: $P \propto \frac{1}{V}$ or $PV = \text{constant}$.
$1$. Graph $(b)$: Represents $PV$ vs $P$. Since $PV$ is constant,the graph is a horizontal line parallel to the $P$-axis.
$2$. Graph $(c)$: Represents $PV$ vs $V$. Since $PV$ is constant,the graph is a horizontal line parallel to the $V$-axis.
Both graphs $(b)$ and $(c)$ correctly represent the constant product of $PV$ as dictated by Boyle's Law.

(C) At higher altitudes,the atmospheric pressure is lower than at sea level.
Because the boiling point of water is dependent on external pressure,a decrease in atmospheric pressure leads to a decrease in the boiling point of water.
Since the water boils at a lower temperature,it cannot provide enough heat to cook the food efficiently,thus taking more time.

(C) According to the ideal gas equation,$PV = nRT$,where $n$ is the number of moles.
Since the number of molecules is directly proportional to the number of moles $(n = \frac{N}{N_A})$,the number of molecules will be equal if the number of moles is equal.
For equal volumes $(V)$ of different gases,the number of moles $(n)$ will be equal if the temperature $(T)$ and pressure $(P)$ of all the flasks are the same.
Therefore,the correct condition is that the temperature and pressure of all the flasks are the same.

(B) For gas $X$: $P_X V = \frac{m_X}{M_X} RT \Rightarrow 2V = \frac{1}{M_X} RT$ ...$(i)$
For gas $Y$: $P_Y V = \frac{m_Y}{M_Y} RT$
Given total pressure $P_{total} = 3 \ atm$,so $P_Y = P_{total} - P_X = 3 - 2 = 1 \ atm$.
Substituting values for gas $Y$: $1 \cdot V = \frac{2}{M_Y} RT$ ...$(ii)$
Dividing equation $(i)$ by $(ii)$: $\frac{2}{1} = \frac{1/M_X}{2/M_Y} = \frac{M_Y}{2M_X}$
$4M_X = M_Y$ or $M_Y = 4M_X$.

(A) The number of moles of $N_2$ is $n_{N_2} = \frac{5 \ g}{28 \ g/mol} \approx 0.1786 \ mol$.
The number of moles of $Ar$ is $n_{Ar} = \frac{6 \ g}{40 \ g/mol} = 0.15 \ mol$.
The mole fraction of $N_2$ is $\chi_{N_2} = \frac{n_{N_2}}{n_{N_2} + n_{Ar}} = \frac{0.1786}{0.1786 + 0.15} = \frac{0.1786}{0.3286} \approx 0.5435$.
The partial pressure of $N_2$ is $p_{N_2} = \chi_{N_2} \times p_{Total} = 0.5435 \times 30 \ bar \approx 16.305 \ bar$.
Rounding to the nearest provided option,the value is $16.36 \ bar$.

(A) According to Graham's law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$:
$\frac{r_A}{r_B} = \sqrt{\frac{M_B}{M_A}}$
Given:
$M_A = 100 \ g/mol$
$M_B = 64 \ g/mol$
$r_A = 12 \times 10^{-3} \ mol/s$
Substituting the values:
$\frac{12 \times 10^{-3}}{r_B} = \sqrt{\frac{64}{100}}$
$\frac{12 \times 10^{-3}}{r_B} = \frac{8}{10} = 0.8$
$r_B = \frac{12 \times 10^{-3}}{0.8} = 15 \times 10^{-3} \ mol/s$

(A) The root mean square speed $(u_{rms})$ is directly proportional to the square root of the absolute temperature $(T)$:
$u_{rms} = \sqrt{\frac{3RT}{M}}$
For the same gas,$u_{rms} \propto \sqrt{T}$.
Initial temperature $T_{1} = 30 + 273 = 303 \ K$.
Final temperature $T_{2} = 930 + 273 = 1203 \ K$.
Taking the ratio:
$\frac{u_{2}}{u_{1}} = \sqrt{\frac{T_{2}}{T_{1}}} = \sqrt{\frac{1203}{303}} = \sqrt{3.97} \approx \sqrt{4} = 2$.
Therefore,$u_{2} = 2u_{1}$.
Thus,the root mean square speed of the gas gets doubled.

(D) The kinetic energy $(KE)$ of an ideal gas is directly proportional to its absolute temperature $(T)$.
For one mole of an ideal gas,the kinetic energy is given by the formula:
$KE = \frac{3}{2} RT$
Since $R$ is the universal gas constant,$KE$ depends only on the absolute temperature $T$.

(D) The total kinetic energy $(KE)$ of an ideal gas is given by the formula: $KE = \frac{3}{2} n R T$.
For the kinetic energies to be equal,we set the expressions for $He$ and $Ar$ equal to each other:
$\frac{3}{2} n_{He} R T_{He} = \frac{3}{2} n_{Ar} R T_{Ar}$.
Canceling the common terms $\frac{3}{2}$ and $R$,we get: $n_{He} \times T_{He} = n_{Ar} \times T_{Ar}$.
Given: $n_{He} = 0.30 \ mol$,$n_{Ar} = 0.40 \ mol$,and $T_{Ar} = 400 \ K$.
Substituting the values: $0.30 \times T_{He} = 0.40 \times 400$.
$T_{He} = \frac{0.40 \times 400}{0.30} = \frac{160}{0.30} = 533.33 \ K \approx 533 \ K$.

(C) The average molecular speed $(v_{avg})$ is given by the formula: $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$.
This shows that $v_{avg} \propto \sqrt{\frac{T}{M}}$.
Comparing the ratios of $\frac{T}{M}$ for each gas:
$A: \frac{500}{32} = 15.625$
$B: \frac{400}{44} \approx 9.09$
$C: \frac{200}{4} = 50.0$
$D: \frac{300}{17} \approx 17.65$
Since the ratio $\frac{T}{M}$ is highest for $He$ $(50.0)$,it will have the greatest average speed.

(C) The deviation from ideal gas behavior is primarily determined by the magnitude of intermolecular forces and the molecular size,represented by the van der Waals constants $a$ and $b$.
$NH_3$ molecules exhibit strong intermolecular forces due to hydrogen bonding,which is significantly stronger than the London dispersion forces present in $He$,$CH_4$,and $H_2$.
Because $NH_3$ has the highest attractive forces (highest $a$ value) among the given options,it deviates the most from ideal gas behavior under the same conditions of temperature and pressure.

(D) Statement $(i)$ is incorrect because for real gases,$Z$ can be equal to $1$ at specific conditions (Boyle temperature).
Statement $(ii)$ is correct because real gases behave ideally at low pressure and high temperature.
Statement $(iii)$ is incorrect because real gases possess intermolecular forces of attraction.
Statement $(iv)$ is incorrect because real gases obey the Van der Waals equation,$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$,not the ideal gas equation $PV = nRT$.
Therefore,statements $(i), (iii),$ and $(iv)$ are incorrect. However,based on the provided options,$(iii)$ and $(iv)$ are clearly incorrect.

(A) Real gases deviate from ideal behavior because of the intermolecular forces and the finite volume of gas molecules.
At $high$ $pressure$,the volume of gas molecules becomes significant compared to the total volume.
At $low$ $temperature$,the kinetic energy of molecules decreases,allowing intermolecular forces (van der Waals forces) to become significant.
Therefore,a gas deviates most from ideal behavior under conditions of $low$ $temperature$ and $high$ $pressure$.

(B) Real gases exhibit ideal gas behavior at high temperature and low pressure.
Under these conditions,the volume $V$ is very large,and the correction terms in the van der Waals equation,specifically $\frac{a}{V^2}$ and $b$,become negligible.
Consequently,the van der Waals equation,$(p + \frac{an^2}{V^2})(V - nb) = nRT$,simplifies to the ideal gas equation,$pV = nRT$.

(A) The ease of liquefaction of a gas depends on the magnitude of intermolecular forces of attraction,which is represented by the van der Waals' constant '$a$'.
Greater the value of '$a$',stronger are the intermolecular forces,and easier it is to liquefy the gas.
Among the given gases,$SO_2$ has the highest value of '$a$' due to its polar nature and strong dipole-dipole interactions.
Additionally,$SO_2$ has a high critical temperature,making it the easiest to liquefy under moderate conditions.

(D) The van der Waals' constant $a$ represents the magnitude of intermolecular attractive forces in a gas.
Greater intermolecular attraction leads to a higher value of $a$.
Among the given options,$NH_3$ molecules exhibit strong hydrogen bonding in addition to dipole-dipole interactions.
$H_2$,$O_2$,and $CH_4$ only exhibit weaker London dispersion forces.
Therefore,$NH_3$ has the strongest intermolecular forces and the highest value of $a$.

(A) The mass of an electron is $m_e = 9.109 \times 10^{-31} \ kg$.
The mass of a proton is $m_p = 1.672 \times 10^{-27} \ kg$.
The mass of a neutron is $m_n = 1.674 \times 10^{-27} \ kg$.
To find the ratio $m_e : m_p : m_n$,we divide each mass by the mass of the electron $(m_e)$:
Ratio $= 1 : \frac{1.672 \times 10^{-27}}{9.109 \times 10^{-31}} : \frac{1.674 \times 10^{-27}}{9.109 \times 10^{-31}}$
Ratio $\approx 1 : 1836.15 : 1838.68$.

(A) For a neutral atom,the number of protons and electrons is equal to the atomic number $(Z)$.
For ${}^{13}_{6}C$,the atomic number $Z = 6$,so the number of protons = $6$ and the number of electrons = $6$.
The number of neutrons is calculated by subtracting the atomic number from the mass number $(A)$:
Number of neutrons = $A - Z = 13 - 6 = 7$.
Therefore,the number of protons,neutrons,and electrons are $6, 7, 6$ respectively.

(C) All the given species are isoelectronic,meaning they all contain $10$ electrons.
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
As the positive charge on the cation increases,the effective nuclear charge increases,which pulls the electrons closer to the nucleus,resulting in a smaller size.
Among the given species,$Si^{4+}$ has the highest atomic number $(Z = 14)$ and the highest positive charge,therefore it has the smallest size.

(C) For isoelectronic species,the ionic radii decrease as the nuclear charge increases.
Since all these ions have the same number of electrons ($18$ electrons),the ionic radius depends on the effective nuclear charge $(Z_{eff})$.
As the atomic number $(Z)$ increases from $S$ $(16)$ to $Ca$ $(20)$,the nuclear charge increases,which pulls the electrons closer to the nucleus.
Therefore,the ionic radii show a significant decrease from $S^{2-}$ to $Ca^{2+}$.

(D) Given that $3.011 \times 10^{22}$ atoms of an element weigh $1.15 \ g$.
We know that $1 \ mol$ of an element contains $N_A$ atoms,where $N_A \approx 6.022 \times 10^{23} \ atoms/mol$.
Mass of $6.022 \times 10^{23}$ atoms (molar mass) $= \frac{1.15 \ g}{3.011 \times 10^{22} \ atoms} \times 6.022 \times 10^{23} \ atoms/mol$.
$= 1.15 \times 20 = 23 \ g/mol$.
Therefore,the atomic mass of the element is $23 \ amu$.

(D) According to Bohr's model,the radius of the $n^{th}$ orbit is given by the formula: $r_n = a_0 \frac{n^2}{Z}$,where $a_0$ is the Bohr radius,$n$ is the principal quantum number,and $Z$ is the atomic number.
For the $1^{st}$ orbit $(n=1)$,$r_1 = a_0 \frac{1^2}{Z} = \frac{a_0}{Z}$.
For the $3^{rd}$ orbit $(n=3)$,$r_3 = a_0 \frac{3^2}{Z} = 9 \times \frac{a_0}{Z}$.
Comparing the two,we find that $r_3 = 9 \times r_1$.
Therefore,the radius of the $3^{rd}$ orbit is $9$ times the radius of the $1^{st}$ orbit.

(A) The spectrum of an atom or ion depends on the number of electrons present in it.
Helium $(He)$ has $2$ electrons.
Among the given options,$Li^{+}$ (Lithium ion) has an atomic number of $3$,so $Li^{+}$ has $3 - 1 = 2$ electrons.
Since both $He$ and $Li^{+}$ have the same number of electrons $(2)$,their spectra are expected to be similar.

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by $E_n = -13.6 \frac{Z^2}{n^2} \ eV$.
For a hydrogen atom $(Z=1)$,the energy difference between two adjacent levels $n$ and $n+1$ is $\Delta E = E_{n+1} - E_n = 13.6 \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right) \ eV$.
As $n$ increases,the term $\frac{1}{n^2}$ decreases rapidly,and the difference between $\frac{1}{n^2}$ and $\frac{1}{(n+1)^2}$ also decreases.
Therefore,the energy difference $\Delta E$ decreases as the Principal Quantum number $n$ increases.

(D) The Rydberg formula for the wavelength of a transition is given by $\frac{1}{\lambda} = R_{H} Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
For the $He^{+}$ ion,$Z=2$. The transition from $n_2=4$ to $n_1=2$ gives:
$\frac{1}{\lambda} = R_{H} (2)^2 \left[ \frac{1}{2^2} - \frac{1}{4^2} \right] = 4 R_{H} \left[ \frac{1}{4} - \frac{1}{16} \right] = 4 R_{H} \left( \frac{3}{16} \right) = \frac{3}{4} R_{H}$.
For the hydrogen atom $(H)$,$Z=1$. We check the transitions:
For option $(d)$,$n_2=2$ to $n_1=1$:
$\frac{1}{\lambda} = R_{H} (1)^2 \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = R_{H} \left( 1 - \frac{1}{4} \right) = \frac{3}{4} R_{H}$.
Since the values match,option $(d)$ is the correct answer.

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6 \ eV}{n^2}$.
For the ground state $(n_1 = 1)$,the energy is $E_1 = -13.6 \ eV$.
When the electron absorbs energy $\Delta E = 12.75 \ eV$,it jumps to a higher orbit $n_2$.
The energy of the final state is $E_{n_2} = E_1 + \Delta E = -13.6 + 12.75 = -0.85 \ eV$.
Using the formula $E_{n_2} = -\frac{13.6}{n_2^2}$,we get:
$-0.85 = -\frac{13.6}{n_2^2}$
$n_2^2 = \frac{13.6}{0.85} = 16$
$n_2 = 4$.
Therefore,the electron jumps to the $4^{th}$ orbit.

(B) The spectral series in the hydrogen spectrum are Lyman $(n_1=1)$,Balmer $(n_1=2)$,Paschen $(n_1=3)$,Brackett $(n_1=4)$,and Pfund $(n_1=5)$.
For the shortest wavelength,the energy transition must be maximum,which occurs for the Lyman series $(n_1=1)$ with $n_2=\infty$.
Using the Rydberg formula: $\frac{1}{\lambda} = R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
Substituting $n_1=1$ and $n_2=\infty$: $\frac{1}{\lambda} = R_H \left[ \frac{1}{1^2} - \frac{1}{\infty^2} \right] = R_H$.
Given $R_H \approx 1.097 \times 10^7 \ m^{-1}$,we get $\lambda = \frac{1}{R_H} \approx 9.117 \times 10^{-8} \ m = 91.2 \ nm$.

(B) According to the Rydberg formula for the hydrogen spectrum: $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Balmer series,$n_1 = 2$.
For the first line,$n_2 = 3$. Given $\lambda_1 = 656 \ nm$,we have $\frac{1}{656} = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = R_H \left( \frac{5}{36} \right) \dots (i)$.
For the second line,$n_2 = 4$. So,$\frac{1}{\lambda_2} = R_H \left( \frac{1}{4} - \frac{1}{16} \right) = R_H \left( \frac{3}{16} \right) \dots (ii)$.
Dividing $(i)$ by $(ii)$,we get $\frac{\lambda_2}{656} = \frac{5/36}{3/16} = \frac{5}{36} \times \frac{16}{3} = \frac{20}{27}$.
$\lambda_2 = 656 \times \frac{20}{27} \approx 485.9 \ nm$.
For the limiting line,$n_2 = \infty$. So,$\frac{1}{\lambda_{\infty}} = R_H \left( \frac{1}{4} - 0 \right) = \frac{R_H}{4}$.
From $(i)$,$R_H = \frac{36}{5 \times 656}$.
$\frac{1}{\lambda_{\infty}} = \frac{36}{5 \times 656 \times 4} = \frac{9}{5 \times 656} = \frac{9}{3280}$.
$\lambda_{\infty} = \frac{3280}{9} \approx 364.4 \ nm$.

(C) According to the energy relation,$E = \frac{hc}{\lambda}$,which implies $E \propto \frac{1}{\lambda}$.
Since the energy of the emitted photon is inversely proportional to its wavelength,the transition with the largest energy difference will result in the shortest (least) wavelength.
The energy difference $\Delta E$ is given by $\Delta E = 13.6 \times Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \text{ eV}$.
Comparing the transitions:
$(A)$ $n_4 \longrightarrow n_3$: $\Delta E \propto (\frac{1}{9} - \frac{1}{16}) = \frac{7}{144} \approx 0.048$
$(B)$ $n_4 \longrightarrow n_2$: $\Delta E \propto (\frac{1}{4} - \frac{1}{16}) = \frac{3}{16} = 0.1875$
$(C)$ $n_4 \longrightarrow n_1$: $\Delta E \propto (\frac{1}{1} - \frac{1}{16}) = \frac{15}{16} = 0.9375$
$(D)$ $n_2 \longrightarrow n_1$: $\Delta E \propto (\frac{1}{1} - \frac{1}{4}) = \frac{3}{4} = 0.75$
The transition $n_4 \longrightarrow n_1$ has the largest energy difference,therefore it corresponds to the least wavelength.

(B) According to the De Broglie relation,the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.
Given that both particles have the same velocity $(v_A = v_B = v)$,the wavelength is inversely proportional to the mass: $\lambda \propto \frac{1}{m}$.
We are given $\lambda_A = 2\lambda_B$.
Substituting the relation: $\frac{h}{m_A v} = 2 \times \frac{h}{m_B v}$.
Simplifying this,we get $\frac{1}{m_A} = \frac{2}{m_B}$,which implies $m_B = 2m_A$ or $m_A = \frac{1}{2}m_B$.
Therefore,the mass of particle $A$ is half that of particle $B$.

(B) According to Planck's quantum theory,the energy of a photon is given by the relation:
$E = \frac{hc}{\lambda}$
This implies that energy is inversely proportional to wavelength:
$E \propto \frac{1}{\lambda} \Rightarrow \lambda \propto \frac{1}{E}$
Given the energies $E_1 = 25 \ eV$ and $E_2 = 100 \ eV$,the ratio of their wavelengths is:
$\frac{\lambda_1}{\lambda_2} = \frac{E_2}{E_1}$
Substituting the given values:
$\frac{\lambda_1}{\lambda_2} = \frac{100 \ eV}{25 \ eV} = \frac{4}{1}$
Therefore,the ratio $\lambda_1: \lambda_2$ is $4:1$.

(D) According to Einstein's photoelectric equation,the kinetic energy $(KE)$ of the ejected photoelectron is given by: $KE = h\nu - \varphi$,where $h\nu$ is the energy of incident radiation and $\varphi$ is the work function of the metal.
From this relation,it is clear that for a constant incident energy $(h\nu)$,the kinetic energy $(KE)$ is inversely proportional to the work function $(\varphi)$ of the metal.
Therefore,the metal with the highest work function will result in the lowest kinetic energy of the ejected photoelectrons.
The order of work functions for the given metals is: $Cu (4.70 \ eV) > Ag (4.26 \ eV) > Li (2.90 \ eV) > Na (2.75 \ eV)$.
Since $Cu$ has the highest work function,it will exhibit the lowest kinetic energy for the ejected photoelectrons.

(C) The work function $\phi$ is given by $\phi = 3.75 \ eV$.
Converting to Joules: $\phi = 3.75 \times 1.602 \times 10^{-19} \ J \approx 6.0075 \times 10^{-19} \ J$.
The threshold wavelength $\lambda_0$ is calculated using the formula $\lambda_0 = \frac{hc}{\phi}$.
Using $h = 6.626 \times 10^{-34} \ J \cdot s$ and $c = 3 \times 10^8 \ m/s$:
$\lambda_0 = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{6.0075 \times 10^{-19}} \approx 3.308 \times 10^{-7} \ m$.
Converting to nanometers: $\lambda_0 \approx 330.8 \ nm$.
Thus,the approximate threshold wavelength is $330 \ nm$.

(D) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$,where $p$ is the momentum.
Since kinetic energy $E = \frac{p^2}{2m}$,we have $p = \sqrt{2mE}$.
Substituting this into the wavelength formula,we get $\lambda = \frac{h}{\sqrt{2mE}}$.
Given that $h$ and $E$ are constants for both particles,$\lambda \propto \frac{1}{\sqrt{m}}$.
For masses $m_1 = m$ and $m_2 = 2m$,the ratio of wavelengths is $\frac{\lambda_1}{\lambda_2} = \frac{\sqrt{m_2}}{\sqrt{m_1}} = \frac{\sqrt{2m}}{\sqrt{m}} = \frac{\sqrt{2}}{1}$.
Thus,the ratio is $\sqrt{2}: 1$.