Mix Test - 3

Given:

\(T_{1}=8 T_{2}\) and \(Z=1\) (for hydrogen)

We know that the time period for the electron moving in the \(n^{\text {th }}\) orbit is given as,

\( T_{n}=\frac{4 \epsilon_{\theta}^{2} n^{3} h^{3}}{m Z^{2} e^{4}}\ldots\ldots\)(i)

Where \(ϵ_o =\) permittivity, \(h =\) Planck's constant, \(m =\) mass of the electron, \(Z =\) atomic number and \(e =\) charge on the electron

By equation (i), the time period of the electron in the hydrogen atom is given as \((Z =1)\),

\( T_{n}=\frac{4 \epsilon_{o}^{2} n^{3} h^{3}}{m \times 1^{2} \times e^{4}}\)

\(\Rightarrow {T}_{{n}} \propto {n}^{3}\quad\ldots\)(ii)

By equation (ii),

\(T_{1} \propto n_{1}^{3}\quad\ldots\)(iii)

\( T_{2} \propto n_{2}^{3}\quad\ldots\)(iv)

Divide equation (iii) and equation (iv),

\( \frac{T_{1}}{T_{2}}=\frac{n_{1}^{3}}{n_{2}^{3}}\)

\(\Rightarrow \frac{n_{1}^{3}}{n_{2}^{3}}=\frac{8 T_{2}}{T_{2}}\)

\(\Rightarrow \frac{n_{1}}{n_{2}}=\frac{2}{1}\)

\(\Rightarrow {n}_{1}=2 {n}_{2}\quad\ldots\)(v)

When \({n}_{2}=2\)

\(\Rightarrow {n}_{1}=2 \times 2\)

\(\Rightarrow {n}_{1}=4\)

Hence, the correct option is (D).

The size of an atom is of the order of \(10^{-10} \mathrm{~m}\).

Rutherford’s Model of the Atom:

He bombarded the beam of alpha particles on a very thin gold foil. While bombarding he observes the number of scattering of particles:

• Most of the particles passed either undeviated or with a very small deviation.
• Some deviated by large angles.
• 1 out of 8000 was deflected by more than 90°.

The conclusion to this deflection:

• Most of the space of an atom is empty.
• At the centre of an atom having a tiny positively charged particle called a nucleus (its size is of approximately \(10^{-15}\) m or 1 fermi).
• The centre nucleus has all the mass of an atom.
• The amount of positive charge at the nucleus is equal to the total amount of negative charges on all the electrons of the atom.
• All the electrons revolve around the nucleus and coulomb force provide the centripetal force.

According to Rutherford’s model Atom consists of a central core which is called the atomic nucleus. Size of the nucleus is about the order of \(10^{-15} \mathrm{~m}\) that is very small. Whereas the size of an atom is of the order of \(10^{-10} \mathrm{~m}\).

Hence, the correct option is (B).

Given:

Half-life = 2 years, total time = 6 years, and let the initial amount of radioactive substance \((N_{0})\)= 1

Since, half-life is 2 years so 6 years is equal to 3 half-lives.

We know that the fraction of substance remains after n half-lives,

\( N^{\prime}=\frac{N_{0}}{2^{n}}\)

Where \(\mathrm{N}'=\) amount remaining after \(\mathrm{n}\) half-lives and \(\mathrm{N}_{0}=\) initial amount

So, \( N^{\prime}=\frac{N_{0}}{2^{3}}\)

\(\Rightarrow N^{\prime}=\frac{N_{0}}{8}=\frac{1}{8}\)

Hence, the correct option is (B)

The spectrum of an oil flame is an example for continuous emission spectrum.

Spectrum: When white light falls on a prism, different wavelengths' waves are deviated in different directions by the prism. The image obtained is coloured images of the slit and this image is called a spectrum. The spectra obtained are classified mainly into two types:

(i) emission spectra and (ii) absorption spectra.

(i) Continuous emission spectrum: It consists of unbroken luminous bands of all wavelengths containing all the colours from violet to red. These spectra depend on the temperature of the source only and are independent of the characteristic of the source.

For example: Incandescent solids, Carbon arc, liquids, electric filament lamps, etc.

(ii) Continuous absorption spectrum: When a pure green glass plate is placed in the path of white light, it absorbs everything except green and so it gives a continuous absorption spectrum.

Since the spectrum of an oil flame consists of continuously varying wavelengths in a definite wavelength range, it is an example of a continuous emission spectrum.

Hence, the correct option is (D).

According to Bohr's atomic model, the speed of the electron in the nth orbit is given as,

\( v_{n}=2.2 \times 10^{6} \frac{Z}{n} \mathrm{~m} / \mathrm{sec}\quad\ldots\)(i)

Where \(\mathrm{Z}=\) atomic number

For ground-state \(n=1\),

\(v_{n}=2.2 \times 10^{6} \times \frac{Z}{1} \mathrm{~m} / \mathrm{sec}\)

\(\Rightarrow {v}_{{n}} \propto \mathrm{Z}\quad\ldots\)(ii)

For \(\mathrm{H}\) atom, \(\mathrm{Z}=1\)

For \(\mathrm{He}^{+}\mathrm{ion}, \mathrm{Z}=2\)

For \(\mathrm{L i}^{+2}\) ion, \(\mathrm{Z}=3\)

By equation (ii) we can say that the speed of the electron in the ground state is directly proportional to the atomic number.

We know that the atomic number of Li⁺² is maximum compared to the H atom and He⁺ ion. So the speed of the electron in the ground state is maximum for Li⁺² ion.

Hence, the correct option is (C).

Given:

Two particles with respective half-lives 1620 and 810 years.

According to the law of Rutherford - Soddy the number of atoms left after n number of Half-life is:

\(N=N_{0}\left(\frac{1}{2}\right)^{n}\). Where \(\mathrm{N}_{0}\) is the original number of atoms. 

The number of half-lives \(n=\frac{\text { time of deccay }}{\text { effective half } \text { life }}\)

The relation between effective disintegration constant \(\lambda\) and half life \(T\) is:

\(\lambda=\frac{\log_{ e}2}{T}\)

Now, \(\lambda_{1}+\lambda_{2}=\frac{\log_{ e}2}{T_{1}}+\frac{\log_{ e}2}{T_{2}}\)

The effective half-life is:

\(\frac{1}{T}=\frac{1}{T_{1}}+\frac{1}{T_{2}} \)

\(=\frac{1}{1620}+\frac{1}{810} \)

\(=\frac{1+2}{1620} \)

\(\Rightarrow T=540\)

Therefore, the number of half-lives is,

\(n=\frac{\text { time of deccay }}{\text { effective half }\operatorname{life}} \)

\(n=\frac{t}{540}\)

\(N=N_{0}\left(\frac{1}{2}\right)^{n}\)

Put the value of \(n\) in above equation:

\(\frac{N}{N_{0}}=\left(\frac{1}{2}\right)^{n}=\left(\frac{1}{2}\right)^{\frac{t}{540}} \)

Now according to the question, 

\(\frac{N}{N_{0}}=\frac{1}{4}\)

\(\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\frac{t}{540}} \)

On comparing the power of both sides,

\(2=\frac{t}{540} \)

\(t=1080\)

Therefore, the time in years, after which one-fourth of the material remains is 1080 years.

Hence, the correct option is (A).

Given:

\(n_{1}\) = 3 and \(n_{2}\) = 4

As we know,

When an electron jumps from \(n_{2}\) orbit to the \(n_{1}\) orbit \((n_{2}>n_{1})\), the energy of the atom changes from \(En_{2}\) to \(En_{1}\). This extra energy \((En_{2} - En_{1})\) is emitted as a photon of electromagnetic radiation. This corresponding wavelength is given as,

\(\frac{1}{\lambda}=R Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)\).......(i)

Where, Rydberg constant \((\mathrm{R})=1.0973 \times 10^{7} \mathrm{~m}^{-1}\) , \(Z=2\), for \(\mathrm{He}^{+}\)ion

Put all the given values in (i),

\(\frac{1}{\lambda}=4 R\left(\frac{1}{9}-\frac{1}{16}\right)\)

\(=4 R \times \frac{7}{144}\)

\(=\frac{7}{36} R\)

\(\Rightarrow\lambda=\frac{36}{7 R}\)

\(=\frac{36}{7 \times 1.0973 \times 10^{7}}\)

\(=468 \mathrm{~nm}\)

Hence, the correct option is (B).

The ionization energy of the hydrogen atom in the excited state means the atom is already in the first excited state and n  = 2.

The energy of the electron in any excited state is given by,

\( E_{n}=-13.6 \frac{Z^{2}}{n^{2}} \mathrm{e V}\)

The above equation can be written for the excited state of the hydrogen atom as,

\( E_{n}=\frac{-13.6}{2^{2}} \mathrm{e V}\)

\(=-3.4 \mathrm{e V}\)

Hence, the correct option is (B).

It represents the Paschen series.

Hydrogen Spectrum and Spectral series: When a hydrogen atom is excited, it returns to its normal unexcited (or ground state) state by emitting the energy it had absorbed earlier.

This energy is given out by the atom in the form of radiations of different wavelengths as the electron jumps down from a higher to a lower orbit. The transition from different orbits causes different wavelengths, these constitute spectral series which are characteristic of the atom emitting them. When observed through a spectroscope, these radiations are imaged as sharp and straight vertical lines of a single colour.

Mainly there are five series and each series is named after it's discovered as Lyman series (n₁ = 1), Balmer series (n₁​ = 2), Paschen series (n₁​ = 3), Bracket series (n₁​ = 4), and Pfund series (n₁​ = 5).

Hence, the correct option is (D).

Given:

\(\mathrm{Z}=1 , \mathrm{n}_{1}=2 , \mathrm{n}_{2}=4 \)

We know that wavelength,

\(\frac{1}{\lambda}=Z^{2} R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \)

\(=R\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right) \)

\(=R\left(\frac{1}{4}-\frac{1}{16}\right) \)

\(\Rightarrow \lambda=\frac{16}{3 R}\)

Hence, the correct option is (A).

Given:

A nucleons with \(\mathrm{z}=92\)

Partical emission Sequence \(\rightarrow 2 \alpha, 2 \beta^{-1}, 4 \alpha, 2 \beta^{-1}, \alpha, 2 \beta^{+}, \alpha\)

Total \(=8 \alpha\) particles, \(4 \beta^{-1}, 2 \beta^{+}\)particles emitted.

we know that:

\(1\alpha\) particle decreases \(\mathrm{z}\) by \(2\). 

\( 1 \beta^{+}\) decay decreases z by \(1\).

\(1 \beta^{-}\)decay increases \(\mathrm{z}\) by \(1\).

\( \mathrm{z}_{\text {final }}=92-8 \alpha_{\text {decay }}-2 \beta_{\text {decay }}^{+}+4\) beta \(^{-}_{decay }\) 

\(=92-16-2+4\) 

\(\mathrm{z}_{\mathrm{final}}=78\)

Hence, the correct option is (B).

Given that transition is from n = 4 to n = 1 (ground state). So transition spectral lines

\(N=\frac{n(n-1)}{2}\)

\(N=\frac{4(4-1)}{2}=6\)

Hence, the correct option is (C).

Bohr model is applicable to both the Hydrogen atom and the He⁺ atom.

Hydrogenic atoms are atoms consisting of a nucleus with positive charge Ze⁺ and a single electron, where Z is the proton number. For examples are a hydrogen atom, singly ionized helium, doubly ionized lithium, and so forth. Bohr model is applicable to hydrogenic atoms. It cannot be extended even to mere two-electron atoms such as helium.

Bohr’s model correctly predicts the frequencies of the light emitted by hydrogenic atoms, the model is unable to explain the relative intensities of the frequencies in the spectrum.

Hence, the correct option is (C).

Given:

Rato of mass numbers \(A_{1}: A_{2}=27: 125\)

\( \frac{A_{1}}{A_{2}}=\frac{27}{125}\)

Radius of nucleus is \(\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{\frac{1 }{ 3}}\)

So,

\(\mathrm{R}_{1}=\mathrm{R}_{0} \mathrm{~A}_{1}^{\frac{1 }{ 3}}\)......(i)

\(\mathrm{R}_{2}=\mathrm{R}_{0} \mathrm{~A}_{2}^{\frac{1 }{ 3}}\).....(ii)

Dividing (i) and (ii),

\(\frac{R_{1}}{R_{2}} =\left(\frac{A_{1}}{A_{2}}\right)^{\frac{1}{ 3}} \)

\(=\left(\frac{27}{125}\right)^{\frac{1 }{ 3}} \)

\(=\frac{3}{5}\)

Thus, Ratio of radii is \(3: 5\).

Hence, the correct option is (A).

When an electron in Hydrogen atom transit from higher energy orbit to 2nd orbit. (outer orbit n₂ > 2 to the orbit n₁ = 2) known as Balmer Series.

When atoms are excited they emit light of certain wavelengths that correspond to different colors. Due to the electron making transitions between two energy levels in an atom. The light emission can be seen as a series of colored lines, known as atomic spectra.

The Balmer series: It includes the lines due to transitions from an outer orbit n₂ > 2 to the orbit n₁ = 2. Four of the Balmer lines lie in the "visible" part of the spectrum. The energy released from this spectrum gives the visible part of the spectrum.

Hence, the correct option is (D).

Alpha particle emission reduces the atomic number by \(2 .\) So, Atomic Number of \(\mathrm{Y}=\) Atomic Number of \(\mathrm{Z}-2\)

Beta emission increases the atomic number by \(1 .\) So, Atomic Number of \(\mathrm{X}\) is that of \(\mathrm{Y}+1\) Similarly, Atomic number of \(\mathrm{R}\) is Atomic Number of \(\mathrm{X}+1\), which is Same as \(\mathrm{Y}+2\) so, \(\mathrm{R}\) has same atomic number as \(\mathrm{Z}\), but different mass numbers. i.e., \(\mathrm{R}\) and \(\mathrm{Z}\) are isotopes.

Hence, the correct option is (B).

Both statements (A) and (B) are correct for the Bohr model.

The Bohr model is applicable to hydrogenic atoms. It cannot be extended even to mere two-electron atoms such as helium.

• The analysis of atoms with more than one electron was attempted on the lines of Bohr’s model for hydrogenic atoms but did not meet with any success.
• The formulation of the Bohr model involves electrical force between the positively charged nucleus and electron. It does not include the electrical forces between electrons which necessarily appear in multi-electron atoms.

Bohr’s model correctly predicts the frequencies of the light emitted by hydrogenic atoms, the model is unable to explain the relative intensities of the frequencies in the spectrum.

Hence, the correct option is (C).

Paschen series: When an electron in a Hydrogen atom transit from higher energy orbit to 3rd orbit. (outer orbit \(n_{2}\) = n > 3 to the orbit \(n_{1}\) = 3) known as Paschen Series.

So the empirical formula for the observed wavelengths (λ) for hydrogen (Z = 1) is,

\(R_{\infty}=1.1\)

\(\frac{1}{\lambda}=R_{\infty}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)\).....(i)

For \(\lambda_{\min },\) \(n_{1}=3\) and \( \mathrm{n}_{2}=\infty\)

Put above values in (i),

\(\frac{1}{\lambda_{\min }}=R_{\infty}\left(\frac{1}{3^{2}}-0\right)\)

\(\Rightarrow \frac{1}{\lambda_{\min }}=1.1 \times 10^{7}\left(\frac{1}{3^{2}}\right)\)

\(\Rightarrow \lambda_{\min }=0.818 \mu \mathrm{m}\)

For \(\lambda_{\max },\)  \(n_{1}=3\) and \( n_{2}=4\)

Put above values in (i),

\(\frac{1}{\lambda_{\max }}=R_{\infty}\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right)\)

\(\Rightarrow \frac{1}{\lambda_{\max }}=1.1 \times 10^{7}\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right)\)

\(\Rightarrow \lambda_{\max }=1.89 \mu \mathrm{m}\)

\(\lambda_{\min }=0.818 \mu \mathrm{m}\) and \(\lambda_{\max }=1.89 \mu \mathrm{m}\)

Hence, the correct option is (C).

We know that, if an electron jumps from higher energy levels n2 = 2,3,4... to n1 =1 then the line of the spectrum is Lyman series.

\( \frac{1}{\lambda}=R\left(\frac{1}{1^{2}}-\frac{1}{n_{2}^{2}}\right)\)

For the longest wavelength of the Lyman series:

\(\frac{1}{\lambda_{l}}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)\)

\(n_{1}=1, n_{2}=2\)

\(\Rightarrow \frac{1}{\lambda_{l}}=R\left(\frac{1}{1}-\frac{1}{4}\right)\)

\(=R  \frac{3}{4}\).....(ii)

For the longest wavelength of the Balmer series:

\(n_{1}=2, n_{2}=3\)

\( \frac{1}{\lambda_{b}}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)\)

\(= R\left(\frac{1}{4}-\frac{1}{9}\right)\)

\(=R \frac{5}{36}\).....(ii)

From (i) and (ii),

\( \frac{\lambda_{l}}{\lambda_{b}}=\frac{\frac{5}{36}}{\frac{3}{4}}\)

\(=\frac{5}{27}\)

Hence, the correct option is (A).

Given:

The electron is in 2nd orbit so \(n=2\).

The energy of an electron at \(\mathrm{n}^{\text {th }}\) orbit is given by:

\( E=-\frac{13.6}{n^{2}} \mathrm{e V}\)

So, the energy of an electron at the \(2^{\text {nd }}\) orbit will be:

\( E=-\frac{13.6}{2^{2}}\)

\(=-3.4 e V\)

Hence, the correct option is (B).

Given:

\(n_{1}=3\) and \(n_{2}=4\)

The magnitude of the electron's angular momentum in 3rd orbit is:

\( L_{1}=\frac{3 h}{2 \pi}\)

The magnitude of the electron's angular momentum in 4 th orbit is:

\( L_{2}=\frac{4 h}{2 \pi}\)

Change in angular momentum

\( \Delta L=L_{2}-L_{1}\)

\(=\frac{h}{2 \pi}(4-3)\)

\(\Rightarrow \Delta L=\frac{6.64 \times 10^{-34}}{2 \times 3.14}(4-3)\)

\(=1.05 \times 10^{-34} J/S\)

Hence, the correct option is (A).

Given:

\(t_{\frac{1 }{ 2}}=1244 ~\mathrm{sec}.\)

For a first order reaction, Specific rate constant ' \(\mathrm{k}\) ' is related to half-life \(\left(\mathrm{t}_{1 / 2}\right)\) as:

\(t_{\frac{1 }{ 2}}=\frac{0.693}{k}\)

\(\Rightarrow k=\frac{0.693}{t_{\frac{1 }{ 2}}}\).....(i)

Put the given value in (i),

\(k=\frac{0.693}{1244}\)

\(=5.6 \times 10^{-4} \mathrm{sec}^{-1}\)

Hence, the correct option is (B).

Lyman series of hydrogen spectrum corresponds to ultraviolet region.

The Lyman series: It includes the lines emitted by transitions of the electron from an outer orbit of quantum number n₂ > 1 to the 1st orbit of quantum number n₁ = 1. All the energy wavelengths in the Lyman series lie in the ultraviolet band.

Hence, the correct option is (D).

The binding energy of a particle can be defined as the minimum energy required to remove nucleons (Proton or neutron) to an infinite distance from the nucleus. It can be expressed as:

\(\Delta \mathrm{E}=\left(\mathrm{Zm}_{\mathrm{p}}+(\mathrm{A}-\mathrm{Z}) \mathrm{m}_{\mathrm{n}}-\mathrm{M}\right) \mathrm{c}^{2} \times 931.5 \mathrm{MeV}\)

This is derived by using Einstein's mass-energy relation and we can consider binding energy per nucleons as an average energy per nucleon needed to separate a nucleus into its individual nucleons.

According to the graph of binding energy per nucleons vs the atomic mass number of atoms, we can see that,

Iron (Fe) is most stable element since binding energy needed to remove 1 nucleon is highest for iron and as we increase atomic mass the binding energy needed to remove 1 nucleon decreases because of coloumbic repulsion between protons inside nucleus which makes it less stable. So among the give option \({ }^{100} \mathrm{Mo}\) is most stable. i.e., among all option atom with an atomic mass number of 100 will have the highest binding energy per nucleon.

Hence, the correct option is (B).

Given:

The masses of isotopes of neon are, \(19.99 \mathrm{u}, 20.99 \mathrm{u}\) and \(21.99 \mathrm{u}\). Relative abundance of the isotopes are \(90.51 \%, 0.27 \%\) and \(9.22 \%\).

Therefore, Average atomic mass of neon is,

Average atomic mass \(=\sum_{\mathrm{i}=1}^{\mathrm{n}}\frac{\left(\operatorname{mass}_{(\mathrm{i})}\right)}{\left(\operatorname{ Abundance}_{(\mathrm{i})}\right)}\)

\(m =\frac{90.51\% \times 19.99+0.27\% \times 20.99+9.22\% \times 21.99}{(90.51\%+0.27\%+9.22\%)}\)

\(m =\frac{90.51 \times 19.99+0.27 \times 20.99+9.22 \times 21.99}{(90.51+0.27+9.22)}\)

\(=\frac{18.9 .29+5.67+202.75}{100}\)

\(=\frac{2017.7}{100}\)

\(=20.17 u \)

Hence, the correct option is (A).

Alpha decay: A nuclear decay process where an unstable nucleus changes to another element by shooting out a particle composed of two protons and two neutrons.

Beta decay: A type of radioactive decay in which a beta particle (fast energetic electron or positron) is emitted from an atomic nucleus, transforming the original nuclide to an isobar of that nuclide.

\(\alpha\)-decay:

\(\qquad{ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X} \quad \rightarrow \quad{ }_{\mathrm{Z}-2}^{\mathrm{A}-4} \mathrm{Y}+\alpha\)

\(\beta \text {-decay}\):

\(\qquad{ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X} \quad \rightarrow \quad{}_{\mathrm{Z}+1}^{\mathrm{~A}} \mathrm{Y}+\beta^{-}\)

Successive emission of \(1 \alpha\) and \(2 \beta\) particles will eventually cause no change in the atomic number of the daughter nuclei.

So, the daughter nuclei is an isotope of the present nuclei.

Hence, the correct option is (B).

As we know, the mean life \((\tau)\) of a radio-active delay is equal to the inverse of the wavelength \((\lambda)\).

\( \tau=\frac{1}{\lambda}\quad\ldots\)(i)

Now as we know that wavelength \((\lambda)\) is inversely proportional to half-life of radio-active sample.

\( \lambda \propto \frac{1}{t_{\frac{1}{2}}}=\frac{k}{{t }_{\frac{1}{2}}}\), where \(\mathrm{k}=\) proportionality constant, \(t_{\frac{1}{2}}=\) half-life of radioactive sample.

And the value of \(\mathrm{k}=\ln 2=0.693\).

\( \lambda=\frac{0.693}{\frac{\tau}{2}}\quad\ldots\)(ii)

Now from equation (i) and (ii) we have,

\( \lambda=\frac{1}{\tau}=\frac{0.693}{t_{\frac{1}{2}}}\)

On simplifying it we get,

\(t_{\frac{1}{2}}=0.693 \tau\)

Hence, the correct option is (A).

Given:

\(n_{1}=4\) and \(n_{2}=5\)

The wavelength of the radiations emitted from the hydrogen atom is given by:

\(\frac{1}{\lambda_{1}}=R\left[\frac{1}{(4)^{2}}-\frac{1}{(5)^{2}}\right]\)

\(= R\left[\frac{1}{16}-\frac{1}{25}\right]=\frac{9 R}{400}\).....(1)

When electrons jump from \(4^{\text {th }}\) to \(1^{\text {st }}\) orbit, then the wavelength of the radiations emitted from the hydrogen atom is given by:

\(\frac{1}{\lambda_{2}}=R\left[\frac{1}{(1)^{2}}-\frac{1}{(4)^{2}}\right]\)

\(= R\left[\frac{1}{1}-\frac{1}{16}\right]=\frac{15 R}{16}\).....(2)

Divide equation (1) and (2), we get,

\(\frac{\lambda_{1}}{\lambda_{2}}=\frac{400 \times 15 R}{9 R \times 16}\)

\(=\frac{125}{3}\)

\(=125 : 3\)

Hence, the correct option is (D).

Given:

\(\mathrm{Be}^{3+} \rightarrow\) The atomic number of beryllium is 4 i.e., \(\mathrm{z}=4\)

\(3^{\text {rd }}\) orbit \((n)=3\)

Radius of orbit is given by:

\(r=\frac{n^{2} h^{2}}{4 \pi^{2} m e^{2}} \times \frac{1}{Z}=0.529 \times \frac{n^{2}}{Z}\)

Put the given values in above formula,

\(=0.529 \times 10^{-10} \times \frac{3^{2}}{4}\)

\(=1.19 \times 10^{-10} \mathrm{~m}\)

\(=0.119 \times\) \(10^{-9} \mathrm{~m} \approx 0.12 \mathrm{~nm}\)

Hence, the correct option is (C).

Given:

Number of scintillations per minute at an angle 60°, \(n_{1}\) = 8100 m

Number of scintillations per minute at an angle 120°, \(n_{2}\) =?

The scattering in the Rutherford's experiment is proportional to \(\cot ^{4} \frac{\phi}{2}\).

\(\Rightarrow  \frac{n_{2}}{n_{1}}=\frac{\frac{\cot ^{4} \phi_{2} }{ 2}}{\frac{\cot ^{4} \phi_{1}}{ 2}}\)

Therefore,

\(\frac{n_{2}}{n_{1}}=\frac{\cot ^{4}\left(\frac{120^{\circ}}{2}\right)}{\cot ^{4}\left(\frac{60^{\circ}}{2}\right)}\)

\(=\frac{\cot ^{4} 60^{\circ}}{\cot ^{4} 30^{\circ}}\)

\(=\left(\frac{\frac{1}{\sqrt{3}}}{\sqrt{3}}\right)^{4}\)

\(=\frac{1}{81}\)

\(n_{2}=\frac{1}{81} \times n_{1}\)

Put the value of \(n_{1}\) in above equation,

\(=\frac{1}{81} \times 8100\)

\(=100\)

Hence, the correct option is (D).