Atoms and Nuclei

To find the maximum wavelength, we use the ionisation energy as the energy of the photon.

$ \begin{aligned} \text { i.e., } \quad E & =\frac{h c}{\lambda} \\ \Rightarrow \quad \lambda & =\frac{h c}{E}=\frac{1240 \mathrm{eV} \cdot \mathrm{~nm}}{13.6 \mathrm{eV}} \\ {[h c=1240 \mathrm{eV} \cdot \mathrm{~nm}] } & \\ & =91.2 \mathrm{~nm}=912\mathop {\rm{A}}\limits^{\rm{o}}\end{aligned} $

Change in mass number

$ \begin{aligned} & =232-208 \\ & =24 \end{aligned} $

∴ Number of $\alpha$-particles $=\frac{24}{4}=6$

Each beta particle has no change in mass number but increases the atomic number by $1\left({ }_{-1}^0 e\right)$.

Initial atomic number of Thorium $(\mathrm{Th})=90$.

Atomic number after emitting 6 alpha particles $=90-(6 \times 2)=90-12=78$

Final atomic number of Lead $(\mathrm{Pb})=82$.

The increase in atomic number from 78 to 82 must be due to beta decay.

Number of beta particles $=$ Final atomic number - Atomic number after alpha decay

$ \begin{aligned} & =82-78 \\ & =4 \end{aligned} $

Momentum is conserved

$ \begin{aligned} & 200 \times V_2=8 V_1 \\ & V_2=\frac{8}{200} V_1=\frac{1}{25} V_1 \end{aligned} $

Energy of two $\alpha$-particles

$ \begin{aligned} & \quad=\frac{1}{2}\left(m_\alpha\right) V_1^2 \times 2=2 E \text { (Given) } \\ & \Rightarrow \quad \frac{1}{2} m_\alpha V_1^2=E \\ & \Rightarrow \quad \frac{1}{2} \times 4 \times V_1^2=E \quad\left(\therefore m_\alpha=4\right) \\ & \text { or } \quad V_1^2=\frac{E}{2} \end{aligned} $

Energy of $\alpha$-particles and daughter nucleus is,

$ \begin{aligned} \mathrm{KE}= & \frac{1}{2}(200) \cdot V_2^2+2\left(\frac{1}{2} 4 \times V_1^2\right) \\ & =\frac{1}{2} \times 200 \times \frac{1}{25 \times 25} \times V_1^2+4 V_1^2 \\ & =\frac{4}{25} V_1^2+4 V_1^2 \\ & =\frac{104}{25} \times V_1^2 \\ & =\frac{104}{25} \times \frac{E}{2}=\frac{52}{25} E . \end{aligned} $

In classical mechanics, the total energy (E) of a particle moving under an inverse square law force (like the electrostatic force between an electron and a nucleus) determines the nature of its orbit:

1. If E < 0 (Negative Total Energy): The particle is in a bound state. Its orbit is closed and elliptical (with a circular orbit being a special case of an ellipse). The particle is trapped by the central force.

2. If E = 0 (Zero Total Energy): The particle is at the boundary between bound and unbound states. Its orbit is open and parabolic. The particle just barely escapes the influence of the central force (it reaches infinity with zero kinetic energy).

3. If E > 0 (Positive Total Energy): The particle is in an unbound state. Its orbit is open and hyperbolic. The particle approaches the central force, is deflected, and then moves away to infinity, never to return. It has enough kinetic energy to overcome the attractive potential and escape.

Given that the total energy of the electron is positive (E > 0), the electron is in an unbound state. This means it will not be trapped in a closed orbit around the nucleus. Instead, it will follow an open trajectory (a hyperbola).

Let's evaluate the options:

• A. electron will revolve in a circular orbit. Circular orbits are closed orbits and correspond to E < 0. Incorrect.

• B. electron will revolve in an elliptical orbit. Elliptical orbits are closed orbits and correspond to E < 0. Incorrect.

• C. electron will not follow a closed orbit. This matches our understanding for E > 0, where the orbit is open (hyperbolic). Correct.

• D. electron will fall into the nucleus. Falling into the nucleus implies a highly negative energy or zero angular momentum, leading to a collision, or continuous energy loss through radiation. A positive energy means the electron has more than enough energy to escape, not fall in. Incorrect.

The final answer is $\boxed{C}$

Percentage of atoms left

$ \begin{aligned} & =100-87.5 \\ & =12.5 \% \\ \text { Using } \frac{N}{N_0} & =\left(\frac{1}{2}\right)^{\frac{t}{T_{1 / 2}}} \\ \frac{125}{100} & =\left(\frac{1}{2}\right)^{\frac{6}{T_{1 / 2}}} \\ \left(\frac{1}{2}\right)^3 & =\left(\frac{1}{2}\right)^{\frac{6}{T_{1 / 2}}} \Rightarrow 3=\frac{6}{T_{1 / 2}} \\ \Rightarrow \quad T_{1 / 2} & =2 \text { day } \end{aligned} $

Now, in 8 days, 4 half-life get completed

⇒ Fraction of atoms left

$ =\frac{100}{2^4 \times 100}=\frac{6.25}{100} $

⇒ Fraction of atoms decay $=\frac{100-6.25}{100}$

$ =\frac{93.75}{100}=\frac{15}{16} $

The mass number of a nucleus is shown by $A$.

The radius of a nucleus depends on the mass number as $R = R_0 A^{1/3}$, where $R_0$ is a constant.

To compare the radii of two nuclei, use the formula:

$\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1 / 3}$

Here, $\frac{A_1}{A_2} = \frac{27}{125}$.

So,

$\frac{R_1}{R_2} = \left(\frac{27}{125}\right)^{1/3} = \frac{3}{5}$

because $27^{1/3} = 3$ and $125^{1/3} = 5$.

The surface area of a sphere is $S = 4\pi R^2$.

The ratio of their surface areas is:

$\frac{S_1}{S_2} = \frac{4\pi R_1^2}{4\pi R_2^2} = \left(\frac{R_1}{R_2}\right)^2 = \left(\frac{3}{5}\right)^2 = \frac{9}{25}$

So, the ratio of their surface areas is $S_1 : S_2 = 9 : 25$.

The range of weak nuclear force is the order of $10^{-16} \mathrm{~m}$. This force is responsible for radioactive decay and acte over very short distances within the nucleus.

Step 1: Find the Total Energy (TE) from the Given Potential Energy (PE)

The potential energy of the electron is given as -6.8 eV.

We use the relation: $ \mathrm{PE} = 2(\mathrm{TE}) $

This means: $ \mathrm{TE} = \frac{\mathrm{PE}}{2} = \frac{-6.8}{2} = -3.4\,\mathrm{eV} $

Step 2: Find the Value of n (Principal Quantum Number)

For a hydrogen atom, the total energy formula is: $ \mathrm{TE} = \frac{-13.6}{n^2} $

Set this equal to the TE we found: $ -3.4 = \frac{-13.6}{n^2} $

Solve for $ n^2 $: $ n^2 = 4 $

So, $ n = 2 $

Step 3: The Electron is in the Second Orbit

Since $ n = 2 $, the electron is in the second Bohr orbit.

Step 4: Use Bohr's Model and de-Broglie Hypothesis

According to Bohr and de-Broglie:

$ 2\pi r_n = n\lambda $

This gives: $ \lambda = \frac{2\pi r_n}{n} \tag{i} $

Step 5: Use the Formula for the Radius of Orbit

Radius of the nth orbit: $ r_n = n^2 r_0 $

For $ n = 2 $ (second orbit): $ r_2 = 2^2 r_0 = 4r_0 $

Step 6: Calculate the de-Broglie Wavelength

Plug $ r_2 $ into the earlier formula:

$ \lambda = \frac{2\pi r_2}{2} = \frac{2\pi}{2} \times 4r_0 = 4\pi r_0 $

Total time, $t=2$ days

$ \begin{aligned} & =2 \times 24 \text { hours } \\ & =48 \text { hours } \end{aligned} $

Number of intervals of 16 hours $=\frac{48}{16}=3$

Decays per interval = 10\%

Remaining fraction $=0.9$

Remaining fraction after 3 intervals

$ \begin{aligned} f^n & =(0.9)^3 \\ & =0.729=72.9 \% \end{aligned} $

Let $X$ and $Y$ be the mass number and atomic number of nucleus $P$.

$ \therefore{ }_Y^X P \longrightarrow Q+\alpha+2 \beta^{-} $

∴ Mass number of $Q=X-4$

Atomic number of $Q=Y-2+2=Y$

Thus, we see that nuclei $P$ and $Q$ have same atomic number but different mass number. Thus, $P$ and $Q$ are isotopes.

Microscopic domain of physics is the phenomenon that deals with the constitution and structure of matter at the minute scales of atoms and nuclei.

$\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

For the second line of Balmer series,

$ \begin{aligned} n_1 & =2 \text { and } n_2=4 \\ \therefore \frac{1}{\lambda_B} & =R\left(\frac{1}{2^2}-\frac{1}{4^2}\right) \\ & =R\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{3 R}{16} \end{aligned} $

The safe level of radiation is 32 times less than what the material gives off now.

Each time a half-life (2.5 hours) passes, the radiation becomes half of what it was before.

We can write this as: $\left(\frac{1}{2}\right)^n = \frac{1}{32}$ where $ n $ is the number of half-lives needed until the radiation is safe.

We know $\frac{1}{32} = \left(\frac{1}{2}\right)^5$. So, $ n = 5 $.

This means it takes 5 half-lives to reach a safe level.

Calculating total time:

$ \frac{t}{T_{1 / 2}} = 5 \Rightarrow t = 5 \times T_{1/2} $

Each half-life is 2.5 hours, so total time is:

$ t = 5 \times 2.5 = 12.5 \text{ hours} $

$ \begin{aligned} &\\ &\text { } \begin{aligned} E & =P \times t \\ = & 64 \times 10^3 \times(60 \times 60) \\ = & 64 \times 10^3 \times 3.6 \times 10^3 \\ = & 2.3 \times 10^8 \mathrm{~J} \end{aligned}\\ &\begin{aligned} & \therefore \text { Energy released per fission }=\frac{E}{N} \\ & \qquad=\frac{2.304 \times 10^8}{7.2 \times 10^{18}}=0.32 \times 10^{-10} \mathrm{~J} \end{aligned} \end{aligned} $

Bose-Einstein statics describes the behaviour of a system of identical, indistinguishable particles with integer spin, known as bosons.

The kinetic energy (KE) of an electron in a hydrogen atom depends on the energy level ($n$) it is in.

Here is the formula for kinetic energy at level $n$:

$ \mathrm{KE}_n = \frac{13.6}{n^2} \ \mathrm{eV} $

Step 1: Identify the energy levels

The "third excited state" means $n = 4$, and the "fourth excited state" means $n = 5$.

Step 2: Calculate the kinetic energy for $n = 4$

$ \mathrm{KE}_3 = \frac{13.6}{4^2} = \frac{13.6}{16} \ \mathrm{eV} $

Step 3: Calculate the kinetic energy for $n = 5$

$ \mathrm{KE}_4 = \frac{13.6}{5^2} = \frac{13.6}{25} \ \mathrm{eV} $

Step 4: Find the ratio of kinetic energies

$ \frac{\mathrm{KE}_3}{\mathrm{KE}_4} = \frac{\frac{13.6}{16}}{\frac{13.6}{25}} = \frac{25}{16} $

So, the ratio of kinetic energies is $25:16$.

In $\beta^{-}$-decay, a neutron transforms into a proton, an electron (which is the $\beta^{-}$particle) and an antinutrino $(\overline{\mathrm{v}})$

Step 1: Formula for Remaining Nuclei

We use the formula $ N = N_0 \left(\frac{1}{2}\right)^n $, where:

• $N$ = number of nuclei left
• $N_0$ = starting number of nuclei
• $n$ = number of half-lives passed

Step 2: Setting Up the Ratio

The question asks for the ratio of nuclei left in $A$ and $B$, so: $ \frac{N_A}{N_B} = \frac{\left( \frac{1}{2} \right)^{n_A}}{\left( \frac{1}{2} \right)^{n_B}} $

Step 3: Simplifying the Ratio

This simplifies to: $ \frac{N_A}{N_B} = \left( \frac{1}{2} \right)^{n_A - n_B} $

Step 4: Finding Number of Half-lives for Each Substance

The number of half-lives $ n $ is found by dividing total time by half-life:

• For $A$: $ n_A = \frac{9}{1.5} = 6 $
• For $B$: $ n_B = \frac{9}{4.5} = 2 $

Step 5: Plugging In the Values

$ \frac{N_A}{N_B} = \left(\frac{1}{2}\right)^{6 - 2} = \left(\frac{1}{2}\right)^4 $

Step 6: Final Answer

$ \left(\frac{1}{2}\right)^4 = \frac{1}{16} $ So, the ratio of the remaining nuclei in $A$ and $B$ after $9$ days is $\frac{1}{16}$.

The frequency for hydrogen spectral lines is found using:

$ f = RC\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) $

Lyman Series (moving to $n_1 = 1$):

First line: Electron goes from $n_2 = 2$ to $n_1 = 1$

$ f_1^{(L)} = RC\left(1 - \frac{1}{4}\right) = RC \frac{3}{4} $

Second line: Electron goes from $n_2 = 3$ to $n_1 = 1$

$ f_2^{(L)} = RC\left(1 - \frac{1}{9}\right) = RC \frac{8}{9} $

Difference between the two Lyman frequencies:

$ \begin{aligned} f &= f_2^{(L)} - f_1^{(L)} \\ &= RC\left(\frac{8}{9} - \frac{3}{4}\right) \\ &= RC\left(\frac{32 - 27}{36}\right) = RC \frac{5}{36} \end{aligned} $

Balmer Series (moving to $n_1 = 2$):

First line: Electron goes from $n_2 = 3$ to $n_1 = 2$

$ f_1^{(B)} = RC\left(\frac{1}{4} - \frac{1}{9}\right) = RC \frac{5}{36} $

Second line: Electron goes from $n_2 = 4$ to $n_1 = 2$

$ f_2^{(B)} = RC\left(\frac{1}{4} - \frac{1}{16}\right) = RC \frac{3}{16} $

Difference between the two Balmer frequencies:

$ \begin{aligned} f' &= f_2^{(B)} - f_1^{(B)} \\ &= RC\left(\frac{3}{16} - \frac{5}{36}\right) \\ &= RC\left(\frac{27 - 20}{144}\right) \\ &= RC\left(\frac{7}{144}\right) \end{aligned} $

Relationship between differences:

$ \begin{aligned} \frac{f'}{f} &= \frac{\left(\frac{7}{144}\right)}{\left(\frac{5}{36}\right)} = \frac{7}{144} \times \frac{36}{5} = \frac{252}{720} = \frac{7}{20} \\ f' &= \frac{7}{20} f \end{aligned} $

Average energy of a neutron emitted in ${ }_{92}^{235} \mathrm{U}$ fission is 2 MeV

$ \begin{aligned} \therefore \quad \varepsilon & =2 \mathrm{MeV} \\ & =2 \times 10^6 \times 1.6 \times 10^{-19} \mathrm{~J} \\ & =3.2 \times 10^{-13} \mathrm{~J}=320 \times 10^{-15} \mathrm{~J} \end{aligned} $

We use this formula to find how much of the substance is left after a certain time:

$ N = N_0 \left(\frac{1}{2}\right)^{t / T_{1/2}} $

It is said that $96.875\%$ of the substance has decayed after $10$ days.

This means that only $3.125\%$ of the substance is left, because $100\% - 96.875\% = 3.125\%$.

We can also write $3.125\%$ as a fraction: $3.125\% = \frac{1}{32}$.

So, $\frac{N}{N_0} = \frac{1}{32}$. We set up the formula:
$ \left(\frac{1}{2}\right)^{t / T_{1/2}} = \frac{1}{32} $
Now, $\frac{1}{32} = \left(\frac{1}{2}\right)^5$, so $t / T_{1/2} = 5$.

So, $t = 5 \times T_{1/2}$.
We know $t = 10$ days. So,
$ T_{1/2} = \frac{10}{5} = 2\ \text{days} $
The half-life of the substance is $2$ days.

In a hydrogen atom, the energy levels are given by

$ E_{n}=\frac{-13.6}{n^{2}} \mathrm{eV} $

where, $ n= $ principal quantum number Energy difference between two energy levels,

$ \begin{array}{l} \Delta E=E_{i}-E_{f} \\\\ h \nu=E_{i}-E_{f} \end{array} $

where, $ E_{i} $ and $ E_{f} $ are the initial and final energy levels.

For the transition from second orbit to first orbit.

$ \begin{array}{l} \Delta E=E_{2}-E_{1} \\\\ h v=\frac{-13.6}{(2)^{2}}-\left(-\frac{13.6}{(1)^{2}}\right) \\\\ h v=-13.6\left[\frac{1}{1}-\frac{1}{4}\right]=-13.6 \times \frac{3}{4} \mathrm{eV} \\\\ v=\frac{-10.2}{h} \mathrm{eV} \end{array} $

Given, $ v=f $, then

$ f=\frac{-10.2}{h} \mathrm{eV} $

For the transition third excited states $ (n=4) $ to first excited state $ (n=2) $.

$ \begin{array}{l} \Delta E^{\prime}=E_{4}-E_{2} \\\\ h f^{\prime}=\frac{-13.6}{4^{2}}+\frac{13.6}{4}=-13.6\left[\frac{1}{4}-\frac{1}{16}\right] \\\\ f^{\prime}=\frac{-2.55}{h} \mathrm{eV} \end{array} $

On comparing Eq. (i) and Eq. (ii), we get

$ f^{\prime}=\frac{f}{4} $

Given,

$ \frac{R_{1}}{R_{2}}=\frac{5}{3} \text { and } A_{2}=27, A_{1}=A $

The radius $ R $ of a nucleus is related to its mass number $ A $ by the formula

$ R \propto A^{\frac{1}{3}} $

Using.Eq. (i),

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

$ \frac{5}{3}=\left(\frac{A}{27}\right)^{\frac{1}{3}} \Rightarrow \frac{5}{3}=\frac{(A)^{\frac{1}{3}}}{3} $

$ A=5^{3} $

$ A=125 $

So, the mass number $ A $ of nucleus $ { }_{52} X^{125} $ is 125 .

Given that the nucleus $ { }_{52} X^{125} $ has 52 protons, the number of neutrons $ N $ in the nucleus is

$ \begin{array}{l} N=A-\text { number of protons } \\\\ N=125-52 \\\\ N=73 \end{array} $

Given,

Half-life of $ A, T_{A}=T $

Half-life of $ B, T_{B}=2 T $

Initial number of nuclei is same for $ A $ and $ B $, it means $ N_{A}=N_{B}=N_{0} $.

The number of remaining nuclei $ N(t) $ after time $ t $ is given by the exponential decay formula,

$ N(t)=N_{0}\left(\frac{1}{2}\right)^{\frac{t}{T_{1} / 2}} $

where, $ N_{0}= $ Initial number of nuclei, $ T_{1 / 2}= $ Half-life After a time $ t=4 T $

$ \begin{aligned} \frac{N_{A}(4 T)}{N_{B}(4 T)} & =\frac{N_{0}\left(\frac{1}{2}\right)^{\frac{4 T}{T}}}{N_{0}\left(\frac{1}{2}\right)^{\frac{4 T}{2 T}}}=\frac{\left(\frac{1}{2}\right)^{4}}{\left(\frac{1}{2}\right)^{2}} =\left(\frac{1}{2}\right)^{2}=\frac{1}{4} \end{aligned} $

Therefore, the ratio of the remaining number of nuclei of $ A $ and $ B $ is $ 1: 4 $.

The ground state energy of a hydrogen atom is given as $-13.6 \, \text{eV}$. The total energy is equivalent to the kinetic energy, while the potential energy is twice the kinetic energy.

Therefore, the potential energy of the hydrogen atom in its ground state is:

$ \text{PE} = 2 \times (-13.6 \, \text{eV}) = -27.2 \, \text{eV} $

For the first excited state of the hydrogen atom, the energy can be calculated using the formula:

$ E_n = -\frac{13.6 \, \text{eV}}{n^2} $

When $n = 2$:

$ E_2 = \frac{-13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} $

Thus, the potential energy, when $n = 2$, is:

$ (\text{PE})_{n=2} = 2 \times (-3.4 \, \text{eV}) = -6.8 \, \text{eV} $

During a beta ($\beta$) decay process, the transformation of the parent nucleus can be represented as follows:

Let the parent nucleus be $ {}_B^A X $.

After a beta ($\beta$) decay, the transformation is:

$ {}_B^A X \longrightarrow {}_{B+1}^A Y + \beta^{-1} $

Here, $ {}_{B+1}^A Y $ represents the daughter nucleus.

In beta decay, the atomic number (denoted by $ B $) increases by 1, but the mass number (denoted by $ A $) remains unchanged. This means that the parent and daughter nuclei have the same mass number but different atomic numbers. Such nuclei are known as isobars.

The decay of a $\mathrm{A}_{92} \mathrm{U}^{238}$ nucleus into a ${ }_{82} \mathrm{Pb}^{214}$ nucleus involves emitting alpha and beta particles. Let's break down this process:

Decay Reaction:

$ { }_{92} \mathrm{U}^{238} \rightarrow { }_{82} \mathrm{~Pb}^{214} $

Mass Number Change:

Initial mass number: 238

Final mass number: 214

Difference in mass number: $238 - 214 = 24$

Each alpha particle reduces the mass number by 4 (since an alpha particle is $ \mathrm{He}^{4}_{2} $). Therefore, the number of alpha particles is:

$ \frac{24}{4} = 6 $

Atomic Number Change:

Initial atomic number: 92

Final atomic number: 82

Difference in atomic number: $92 - 82 = 10$

Alpha Particle Contribution:

Every alpha particle emission decreases the atomic number by 2.

With 6 alpha particles emitted, the atomic number reduces by:

$ 6 \times 2 = 12 $

This brings the atomic number to $92 - 12 = 80$.

Beta Particle Emission:

To adjust back from atomic number 80 to 82, beta particles are necessary.

Each beta particle ($\beta^-$) increases the atomic number by 1.

Therefore, 2 beta particles are emitted to increase the atomic number from 80 to 82.

In conclusion, in the decay process of ${ }_{92} \mathrm{U}^{238}$ to $ { }_{82} \mathrm{~Pb}^{214} $, 6 alpha particles and 2 beta particles are emitted:

$ { }_{92} \mathrm{U}^{238} \rightarrow { }_{82} \mathrm{~Pb}^{214} + 6 \, \alpha + 2 \, \beta^- $

Thus, the numbers of alpha and beta particles emitted are 6 and 2, respectively.

According to Bohr's postulate, the angular momentum is,

$ L=\frac{n h}{2 \pi} $

We have been asked for change in angular momentum in two consecutive orbit.

Let $ n_{1}=n $ and $ n_{2}=n+1 $

$ \begin{aligned} L_{1} & =\frac{n h}{2 \pi} \text { and } L_{2}=\frac{(n+1) h}{2 \pi} \\\\ \Delta L & =L_{2}-L_{1} \\\\ & =\frac{(n+1) h}{2 \pi}-\frac{n h}{2 \pi} \\\\ \Delta L & =\frac{h}{2 \pi} \end{aligned} $