Atoms and Nuclei

The number of moles (n) is given by the formula :

$ \mathrm{n}=\frac{\text { Given Mass }}{\text { Molar Mass }} $

Given :

• Mass of Uranium $=47 \mathrm{~g}$

• Molar Mass of $\mathrm{U}_{92}^{235}=235 \mathrm{~g} / \mathrm{mol}$

$ \mathrm{n}=\frac{47}{235}=\frac{1}{5}=0.2 \mathrm{moles} $

Each mole contains a number of atoms equal to Avogadro's Number ( $N_A$ ).

$ \mathrm{N}=\mathrm{n} \times \mathrm{N}_{\mathrm{A}} $

$\Rightarrow $ $\mathrm{N}=0.2 \times\left(6 \times 10^{23}\right)$

Since each atom (nucleus) releases 190 MeV upon fission :

$ \mathrm{E}=\mathrm{N} \times \text { Energy per fission } $

$\Rightarrow $ $\mathrm{E}=\left(1.2 \times 10^{23}\right) \times 190 \mathrm{MeV}$

$\Rightarrow $ $\mathrm{E}=228 \times 10^{23} \mathrm{MeV}$

$\Rightarrow \alpha \times 10^{23} \mathrm{MeV}=228 \times 10^{23} \mathrm{MeV}$

$ \Rightarrow \alpha=228 $

Therefore, the correct answer is $\mathbf{2 2 8}$.

The given reaction is the proton bombardment of lithium:

$ {}_3^7Li + {}_1^1p \rightarrow 2\, {}_2^4He $

Mass of reactants:

$ m_{\text{reactants}} = m({}_3^7Li) + m({}_1^1p) = 7.0183 + 1.008 = 8.0263~u $

Mass of products:

$ m_{\text{products}} = 2 \times m({}_2^4He) = 2 \times 4.004 = 8.008~u $

Mass defect:

$ \Delta m = m_{\text{reactants}} - m_{\text{products}} = 8.0263 - 8.008 = 0.0183~u $

$ E_{\text{per reaction}} = \Delta m \times 931~\text{MeV} $

$ E_{\text{per reaction}} = 0.0183 \times 931 = 17.04~\text{MeV} $

Convert this to eV:

$ E_{\text{per reaction}} = 17.04 \times 10^6~\text{eV} = 1.704 \times 10^7~\text{eV} $

Given mass of Li:

$ m_{\text{given}} = \frac{7}{17.13}~\text{kg} $

Convert to grams:

$ m_{\text{given}} = \frac{7}{17.13} \times 1000 = 408.6~\text{g (approximately)} $

Molar mass of Li-7 = 7 g/mol.

Number of moles:

$ n = \frac{408.6}{7} = 58.37~\text{mol} $

Number of atoms:

$ N = n \times N_A = 58.37 \times 6.0 \times 10^{23} = 3.50 \times 10^{25} $

Each Li atom undergoes one reaction, so

$ E_{\text{total}} = N \times E_{\text{per reaction}} $

$ E_{\text{total}} = 3.50 \times 10^{25} \times 1.704 \times 10^7 $

$ E_{\text{total}} = 5.96 \times 10^{32}~\text{eV} $

The question states:

$ E_{\text{total}} = \alpha \times 10^{32}~\text{eV} $

$ \Rightarrow \alpha = 5.96 \approx 6 $

Final Answer

$ \boxed{\alpha = 6} $

For an electron moving in a circular orbit, the magnetic field at the centre of the orbit is

$ B=\frac{\mu_0 I}{2r} $

where $I$ is the current due to revolving electron.

Since current is

$ I=\frac{e}{T} $

and the time period

$ T=\frac{2\pi r}{v} $

so,

$ I=\frac{ev}{2\pi r} $

Substituting in the formula of magnetic field,

$ B=\frac{\mu_0}{2r}\cdot \frac{ev}{2\pi r} =\frac{\mu_0 ev}{4\pi r^2} $

So,

$ B \propto \frac{v}{r^2} $

Now from Bohr’s model for hydrogen atom:

$ r_n \propto n^2 \quad \text{and} \quad v_n \propto \frac{1}{n} $

Therefore,

$ B_n \propto \frac{1/n}{(n^2)^2} =\frac{1}{n^5} $

Hence,

$ \frac{B_2}{B_4}=\frac{1/2^5}{1/4^5}=\frac{4^5}{2^5}=\left(\frac{4}{2}\right)^5=2^5=32 $

So, the required ratio is

$ B_2 : B_4 = 32 : 1 $

To find the integer value of $ n $ for which an electron transitions from the fourth excited state in a hydrogen atom, thus emitting a photon with an energy of 2.86 eV, we can follow these steps:

We use the formula for the energy difference associated with electron transitions in a hydrogen atom:

$ E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) $

However, in this question, it seems there's a typo in the original explanation, as both indices are given as $\mathrm{n}_1$. To correct it, we should use this formula:

$ E = 13.6 \left( \frac{1}{n^2} - \frac{1}{5^2} \right) $

where $ n_1 = 5 $ (the fifth energy level or fourth excited state) and $ n_2 = n $ (the state to which the electron transitions).

Given the photon's energy is 2.86 eV, set up the equation:

$ 2.86 = 13.6 \left( \frac{1}{n^2} - \frac{1}{25} \right) $

Solve for $\frac{1}{n^2}$:

$ \frac{1}{n^2} = \frac{2.86}{13.6} + \frac{1}{25} $

Calculate the value:

$ \frac{1}{n^2} = 0.21 + 0.04 = 0.25 $

Find $ n^2 $:

$ n^2 = \frac{1}{0.25} = 4 $

Consequently:

$ n = \sqrt{4} = 2 $

Thus, the electron transitions to the $ n = 2 $ energy state.

To determine the rate of converting ${ }^4 \mathrm{He}$ to ${ }^{12} \mathrm{C}$, we need to calculate the energy released per reaction and use the given power production of the star to find the rate of reactions. The relevant nuclear reaction is:

${ }^4 \mathrm{He} + { }^4 \mathrm{He} + { }^4 \mathrm{He} \rightarrow { }^{12} \mathrm{C} + \mathrm{Q}$

The masses involved in the reaction are given:

• Mass of ${ }^4 \mathrm{He} = 4.0026 \, \mathrm{u}$
• Mass of ${ }^{12} \mathrm{C} = 12 \, \mathrm{u}$

First, we calculate the mass defect (difference between the mass of reactants and products) which will give us the energy released in each reaction:

Mass of reactants: $3 \times 4.0026 \, \mathrm{u} = 12.0078 \, \mathrm{u}$

Mass of product: $12 \, \mathrm{u}$

Mass defect: $12.0078 \, \mathrm{u} - 12 \, \mathrm{u} = 0.0078 \, \mathrm{u}$

We use Einstein's mass-energy equivalence principle, $E = mc^2$, to find the energy released per reaction. The conversion factor between atomic mass units and energy is $1 \, \mathrm{u} = 931.5 \, \mathrm{MeV}$.

Energy released per reaction: $0.0078 \, \mathrm{u} \times 931.5 \, \mathrm{MeV/u} = 7.2627 \, \mathrm{MeV}$

We convert this energy into joules. $1 \, \mathrm{MeV} = 1.60218 \times 10^{-13} \, \mathrm{J}$:

Energy per reaction: $7.2627 \, \mathrm{MeV} \times 1.60218 \times 10^{-13} \, \mathrm{J/MeV} = 1.163 \times 10^{-12} \, \mathrm{J}$

The power generated by the star is given as $5.808 \times 10^{30} \, \mathrm{W}$. The rate of the reaction is the power divided by the energy per reaction:

$\text{Rate of reactions} = \frac{\text{Power}}{\text{Energy per reaction}}$

$ \text{Rate} = \frac{5.808 \times 10^{30} \, \mathrm{W}}{1.163 \times 10^{-12} \, \mathrm{J}}$

$ \text{Rate} = 4.99 \times 10^{42} \, \mathrm{s^{-1}} \simeq 5 \times 10^{42} \mathrm{~s}^{-1}$

Thus, the rate of converting ${ }^4 \mathrm{He}$ to ${ }^{12} \mathrm{C}$ is:

$ \mathrm{n} = 5$

$\begin{aligned} & \frac{1}{2} m v_0^2=\frac{1}{4 \pi \epsilon_0} \frac{z(e)}{r} \\ & \frac{1}{2} \times 6.72 \times 10^{-27} v_0^2=9 \times 10^9 \times \frac{80 \times 2 \times 1.6 \times 1.6 \times 10^{-19} \times 10^{-19}}{4.5 \times 10^{-14}} \\ & v_0^2=\frac{2 \times 9 \times 80 \times 1.6 \times 1.6 \times 2}{6.72 \times 4.5} \times 10^{-38+14+27+9} \\ & v_0^2=243.8 \times 10^{12} \\ & v_0=15.6 \times 10^6 \\ & v_0=156 \times 10^5 \end{aligned}$

Let's approach this problem by understanding the basics and applying the Bohr model to find the energy levels of a hydrogen atom.

The energy of an electron in a hydrogen atom for any given orbit can be defined using the formula:

$E_n = \frac{E}{n^2}$

where:

• $E$ is the energy of the electron in the ground state ($n=1$), and
• $n$ is the principal quantum number (orbit number).

The radius of an orbit in the hydrogen atom, according to the Bohr model, is given by:

$r_n = n^2 a_0$

where:

• $r_n$ is the radius of the nth orbit,
• $a_0$ is the Bohr radius ($0.529$ angstroms or $\mathop A\limits^o$), and
• $n$ is the principal quantum number (orbit number).

Given that:

• The radius of a certain orbit of the hydrogen atom is $8.48 \mathop A\limits^o$, and
• The energy of an electron in this orbit is $E/x$,
• We need to find $x$.

First, let's find $n$, the principal quantum number for the orbit with radius $8.48$ angstroms:

$8.48 = n^2 \times 0.529$

Solving for $n^2$:

$n^2 = \frac{8.48}{0.529}$

$n^2 \approx 16.03$

For simplicity and practicality in the quantum model, $n^2$ approximately equal to 16 would imply $n = 4$, considering $n$ must be a whole number and $16.03$ is close to $16$, which is a perfect square of $4$.

Now, to find $x$, we'll use the energy relationship. Since the energy levels of the hydrogen atom are inversely proportional to the square of the principal quantum number $n$,

$E_{orbit} = \frac{E}{n^2} = \frac{E}{4^2} = \frac{E}{16}$

According to the given information, $E_{orbit} = \frac{E}{x}$, which means:

$\frac{E}{x} = \frac{E}{16}$

Hence,

$x = 16$

$\frac{1}{915}=R_H\left(\frac{1}{1^2}-\frac{1}{\infty^2}\right)\quad$ (For Lyman)

$\Rightarrow \frac{1}{\lambda}=R_H\left(\frac{1}{2^2}-\frac{1}{3^2}\right)\quad$ (For Balmer)

$\Rightarrow \lambda=6588$ $\mathop A\limits^o $

To find the energy released when three helium nuclei combine to form a carbon nucleus, we first need to understand that this process is essentially nuclear fusion, forming a heavier nucleus from lighter ones. The mass defect in this fusion process is the key to calculating the energy released, according to Einstein's equation $E = \Delta mc^2$, where $E$ is the energy released, $\Delta m$ is the mass defect, and $c$ is the speed of light.

The atomic mass of a helium nucleus (also known as an alpha particle) is $4.002603 \, \text{u}$.

1. Calculate the total initial mass of three helium nuclei:

$\text{Total initial mass} = 3 \times 4.002603 \, \text{u} = 12.007809 \, \text{u}$

2. The atomic mass of a carbon nucleus formed by the fusion of three helium nuclei is not directly given, but we can infer it's approximately $12 \, \text{u}$, based on knowledge of isotopes and considering that the question appears to simplify the carbon nucleus to a mass number of 12 (common carbon-12 isotope).

3. Calculate the mass defect ($\Delta m$):

$\Delta m = \text{Total initial mass} - \text{Final mass}$

$\Delta m = 12.007809 \, \text{u} - 12 \, \text{u} = 0.007809 \, \text{u}$

4. Convert the mass defect to energy. Given $1 \, \text{u} = 931 \, \text{MeV/c}^2$, the energy released is calculated using the formula $E = \Delta mc^2$:

$E = 0.007809 \, \text{u} \times 931 \, \text{MeV/u} = 7.271839 \, \text{MeV}$

Since the question asks for the answer in the format of $\times 10^{-2} \, \text{MeV}$, we convert the energy released:

$7.271839 \, \text{MeV} = 727.1839 \times 10^{-2} \, \text{MeV}$

Therefore, the energy released in this reaction is approximately $727.1839 \times 10^{-2} \, \text{MeV}$. The exact value might differ slightly depending on how the atomic mass of the carbon nucleus is considered or rounded in specific scenarios, but based on the information provided, this is a suitable approximation.

Q. value

$\begin{aligned} & =\{(235.0439)-[39.9054+93.9063+1.0086]\} \times 931 \mathrm{~MeV} \\ & =208 \mathrm{~MeV} \end{aligned}$

$\begin{aligned} \Delta E & =13.6 \mathrm{eV}\left(\frac{1}{4}-\frac{1}{9}\right) \\ & =\frac{68}{36} \mathrm{eV}=1.89 \mathrm{eV} \end{aligned}$

Due to recoil of hydrogen atom, the energy of emitted photon will decrease by very small amount.

So for approximate calculations,

$\begin{aligned} & \% \text { charge }= \frac{\Delta E_{\text {atom }}}{\Delta E} \times 100 \\ &=\frac{\frac{\left(\frac{\Delta E}{C}\right)^2}{2 m}}{\Delta E} \times 100 \\ &=\frac{\Delta E}{C^2 \times 2 m} \times 100 \\ &=\frac{1.89 \times 1.6 \times 10^{-19} \times 100}{\left(3 \times 10^8\right)^2 \times 2 \times 1.6 \times 10^{-27}} \\ &=1.05 \times 10^{-7} \% \\ & \therefore n=7 \end{aligned}$

The emission frequency of radiation from a hydrogen-like ion during electron transitions can be understood using the formula derived from Rydberg's equation for hydrogen-like atoms, which is given as:

$E = \frac{E_0}{h} \left( \frac{1}{n^2_1} - \frac{1}{n^2_2} \right)$

where:

• $E$ is the energy of the emitted photon,
• $E_0$ is the Rydberg constant for hydrogen,
• $h$ is Planck's constant,
• $n_1$ and $n_2$ are the principal quantum numbers for the initial and final energy levels, respectively, with $n_2 < n_1$.

Given a transition from $n=2$ to $n=1$ emits radiation with a frequency of $3 \times 10^{15} \mathrm{~Hz}$, we can write:

$\nu_1 = 3 \times 10^{15} \mathrm{~Hz}$

For the transition from $n=3$ to $n=1$, we can use the same principle to find the frequency of the emitted radiation, $\nu_2$, which will depend on the difference in energy levels involved in the transition. The frequency is directly proportional to this energy difference, so we can compare the two transitions using their respective frequencies:

$\frac{\nu_2}{\nu_1} = \frac{\left( \frac{1}{n_{1f}^2} - \frac{1}{n_{1i}^2} \right)}{\left( \frac{1}{n_{2f}^2} - \frac{1}{n_{2i}^2} \right)} = \frac{\left( \frac{1}{1^2} - \frac{1}{3^2} \right)}{\left( \frac{1}{1^2} - \frac{1}{2^2} \right)} = \frac{\left(1 - \frac{1}{9}\right)}{\left(1 - \frac{1}{4}\right)}$

After simplification:

$\frac{\nu_2}{3 \times 10^{15}} = \frac{\left(\frac{8}{9}\right)}{\left(\frac{3}{4}\right)} = \frac{32}{27}$

Thus, to find the frequency $ u_2$ for the transition from $n=3$ to $n=1$:

$\nu_2 = \frac{32}{9} \times 10^{15} \mathrm{Hz}$

Hence, in the given formula $\frac{x}{9} \times 10^{15} \mathrm{~Hz}$ for the frequency, $x$ is equal to $32$.

According to the empirical formula relating the radius of a nucleus ($ R $) with its mass number ($ A $), we know that the radius of a nucleus is proportional to the cube root of its mass number. This relationship is given as:

$ R = R_0 A^{1/3} $

where $ R_0 $ is a constant with an approximate value of 1.2 fermis.

Given that for a nucleus with a mass number 64 has a radius of 4.8 fermis, we can write:

$ 4.8 \text{ fermi} = R_0 \times 64^{1/3} $

Now another nucleus has a radius of 4 fermis:

$ 4 \text{ fermi} = R_0 \times A'^{1/3} $

Where $ A' $ is the mass number of the other nucleus.

Let's solve for $ R_0 $ from the first equation:

$ R_0 = \frac{4.8 \text{ fermi}}{64^{1/3}} $

Now we're going to find the mass number $ A' $ using the second equation and substituting $ R_0 $ from the above:

$ 4 \text{ fermi} = \left( \frac{4.8 \text{ fermi}}{64^{1/3}} \right) \times A'^{1/3} $

Now, we want to find $ A' $ in terms of $ x $ as given by the equation in the question:

$ A' = \frac{1000}{x} $

Substitute $ A' $ in the equation above, we get:

$ 4 = \left( \frac{4.8}{64^{1/3}} \right) \times \left( \frac{1000}{x} \right)^{1/3} $

Let's solve for $ x $:

$ (4)^3 = \left( \frac{4.8}{64^{1/3}} \right)^3 \times \frac{1000}{x} $

$ 4^3 \times x = \left( \frac{4.8}{64^{1/3}} \right)^3 \times 1000 $

$ x = \left( \frac{\left( \frac{4.8}{64^{1/3}} \right)^3 \times 1000}{4^3} \right) $

Now calculate the values:

$ x = \left( \frac{\left( \frac{4.8}{4} \right)^3 \times 1000}{64} \right) $

$ x = \left( \frac{1.2^3 \times 1000}{64} \right) $

$ x = \left( \frac{1.728 \times 1000}{64} \right) $

$ x = \left( \frac{1728}{64} \right) $

$ x = 27 $

Therefore, the value of $ x $ is 27.

For a nucleus

Volume: $\mathrm{V}=\frac{4}{3} \pi \mathrm{R}^3$

$\begin{aligned} & \mathrm{R}=\mathrm{R}_0(\mathrm{A})^{1 / 3} \\ & \mathrm{~V}=\frac{4}{3} \pi \mathrm{R}_0^3 \mathrm{A} \\ & \Rightarrow \frac{\mathrm{V}_2}{\mathrm{~V}_1}=\frac{\mathrm{A}_2}{\mathrm{~A}_1}=4 \end{aligned}$

$\begin{aligned} & \mathrm{E}=\Delta \mathrm{mc}^2 \\ & =0.4 \times 10^{-3} \times\left(3 \times 10^8\right)^2 \\ & =3600 \times 10^7 \mathrm{kWs} \\ & =\frac{3600 \times 10^7}{3600} \mathrm{kWh}=1 \times 10^7 \mathrm{kWh} \end{aligned}$

$\begin{aligned} & E_n=-\frac{13.6}{n^2}=-0.85 \\ & \Rightarrow n=4 \end{aligned}$

No of transition

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

For minimum potential difference electron has to make transition from $n=3$ to $n=2$ state but first electron has to reach to $\mathrm{n}=3$ state from ground state. So, energy of bombarding electron should be equal to energy difference of $\mathrm{n=3}$ and $\mathrm{n=1}$ state.

$\begin{aligned} & \Delta \mathrm{E}=13.6\left[1-\frac{1}{3^2}\right] \mathrm{e}=\mathrm{eV} \\ & \frac{13.6 \times 8}{9}=\mathrm{V} \\ & \mathrm{V}=12.09 \mathrm{~V} \approx 12.1 \mathrm{~V} \end{aligned}$

So, $\alpha=121$

$\begin{aligned} & \Delta \mathbf{E}=\mathbf{1 0 . 2} \mathrm{eV} \\ & \text { Recoil speed }(\mathrm{v})=\frac{\Delta \mathrm{E}}{\mathrm{mc}} \\ & =\frac{10.2 \mathrm{eV}}{1.6 \times 10^{-27} \times 3 \times 10^8} \\ & =\frac{10.2 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-27} \times 3 \times 10^8} \\ & \mathrm{v}=3.4 \mathrm{~m} / \mathrm{s}=\frac{17}{5} \mathrm{~m} / \mathrm{s} \end{aligned}$

Therefore, $x=17$

Longest wavelength corresponds to transition between $\mathrm{n}=3$ and $\mathrm{n}=4$

$\begin{aligned} & \frac{1}{\lambda}=\mathrm{RZ}^2\left(\frac{1}{3^2}-\frac{1}{4^2}\right)=\mathrm{RZ}^2\left(\frac{1}{9}-\frac{1}{16}\right) \\ & =\frac{7 \mathrm{RZ}^2}{9 \times 16} \\ & \Rightarrow \lambda=\frac{144}{7 \mathrm{R}} \text { for } \mathrm{Z}=1 \quad \therefore \alpha=144 \end{aligned}$

To determine the energy released in a nuclear fission process, we use the difference in binding energy (BE) before and after the fission. The formula for energy released ($Q$ value) in the process is given by:

$Q = (\text{Total BE of products}) - (\text{Total BE of reactants})$

In this case, the reactant is a high mass nuclide with atomic mass $A \approx 236$ and a binding energy of $7.6 \mathrm{MeV}/\mathrm{nucleon}$. Each of the two middle mass nuclides formed as products has atomic mass $A \approx 118$ and a binding energy of $8.6 \mathrm{MeV}/\mathrm{nucleon}$.

Therefore, we calculate the total binding energy of reactant and products as follows:

For reactant:

$\text{BE}_{\text{reactant}} = 236 \times 7.6 \mathrm{MeV}$

For products (since there are two identical products):

$\text{BE}_{\text{products}} = 2 \times (118 \times 8.6) \mathrm{MeV}$

Thus, the energy released ($Q$ value) is:

$Q = \text{BE}_{\text{products}} - \text{BE}_{\text{reactant}}$

$Q = 2(118 \times 8.6) - (236 \times 7.6)$

$Q = 236 \times (8.6 - 7.6)$

$Q = 236 \times 1$

$Q = 236 \mathrm{MeV}$

This calculation demonstrates how the difference in binding energy per nucleon before and after fission leads to the release of energy, consistent with the mass-energy equivalence principle.

To find the value of $x$, we need to first determine the expressions for the radii of the specified orbits for $\mathrm{He}^{+}$ and $\mathrm{Be}^{3+}$ according to Bohr's model. The radius of an orbit in a hydrogen-like atom (an atom with only one electron) is given by:

Where:

• $r_n$ is the radius of the nth orbit
• $n$ is the principal quantum number (orbit number)
• $h$ is the Planck's constant
• $\epsilon_0$ is the vacuum permittivity
• $Z$ is the atomic number (number of protons in the nucleus)
• $e$ is the elementary charge
• $m_e$ is the mass of the electron
• $\pi$ is the mathematical constant pi

In this problem, we are looking at the 2nd orbit of $\mathrm{He}^{+}$ (which has an atomic number $Z = 2$) and the 4th orbit of $\mathrm{Be}^{3+}$ (which has an atomic number $Z = 4$). Let's calculate the radii for these orbits:

For the 2nd orbit of $\mathrm{He}^{+}$ ($n_1 = 2$ and $Z_1 = 2$):

For the 4th orbit of $\mathrm{Be}^{3+}$ ($n_2 = 4$ and $Z_2 = 4$):

We are asked to find the ratio $\frac{r_{2}}{r_{1}}$, which is equal to $x: 1$:

By simplifying the expression, we get:

Now we can calculate the value of $x$:

Therefore, the value of $x$ in the ratio $\frac{r_{2}}{r_{1}} = x: 1$ is $\boxed{2}$.

The energy difference formula for transitions between energy levels in a hydrogen atom, which is given by

$ \Delta E = -13.6 \, \text{eV} \times \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) $

where ($n_1$) and ($n_2$) are the initial and final energy levels, respectively. For the Lyman series, the electron transitions to the ground state (($n_1$ = 1)), and for the Balmer series, the electron transitions to the first excited state (($n_1$ = 2)).

From the given information, we have the lowest wavelength in the Lyman series, ($\lambda_1 = 917 \, \text{Å}$). Therefore, the energy difference ($\Delta E$) for the Lyman series is

$ \Delta E = \frac{hc}{\lambda_1} $

where (h) is Planck's constant and (c) is the speed of light.

Similarly, for the Balmer series, the energy difference ($\Delta E$) is

$ \Delta E = -13.6 \, \text{eV} \times \left(\frac{1}{2^2} - \frac{1}{\infty^2}\right) = -13.6 \, \text{eV} \times \frac{1}{4} $

The corresponding wavelength ($\lambda_2$) is

$ \lambda_2 = \frac{hc}{\Delta E} $

By comparing ($\Delta E$) for the Lyman and Balmer series, you found that

$ \frac{\lambda_1}{\lambda_2} = \frac{\Delta E_2}{\Delta E_1} = \frac{1}{4} $

Therefore, the lowest wavelength in the Balmer series is

$ \lambda_2 = 4 \lambda_1 = 4 \times 917 \, \text{Å} = 3668 \, \text{Å} $

The activity of a radioactive substance is defined as the rate of decay or disintegration of the substance. It is given by the following formula:

$A = \lambda N$

where $A$ is the activity, $\lambda$ is the decay constant, and $N$ is the number of radioactive atoms present.

We can use this formula to find the activity of 1.0 $\mu$g (or $10^{-6}$ g) of the substance. First, we need to find the number of radioactive atoms present in 1.0 $\mu$g of the substance. We can use the following formula to do this:

$N = \frac{m}{M} N_A$

where $m$ is the mass of the substance, $M$ is its molar mass, and $N_A$ is Avogadro's number.

Substituting the given values, we get:

$N = \frac{1.0 \times 10^{-6} \, \text{g}}{60 \, \text{g/mol}} \times 6 \times 10^{23} \approx 10^{16} \text{ atoms}$

Now, we can use the formula for activity:

$A = \lambda N = (1.5 \times 10^{-5} \, \text{s}^{-1}) (10^{16}) = 1.5 \times 10^{11} \, \text{Bq}$

Therefore, the activity of 1.0 $\mu$g of the substance is 15 $\times$ 10$^{10}$ Bq.

The Balmer series corresponds to electronic transitions in a hydrogen atom that terminate in the second (n=2) energy level. The spectral lines in the Balmer series are often labeled according to a Greek letter scheme, with H$_\alpha$ corresponding to the n=3 to n=2 transition, H$_\beta$ corresponding to the n=4 to n=2 transition, and so on.

The wavelength of a spectral line in the Balmer series can be calculated using the Rydberg formula:

$ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{n^2} \right) $

where $R_H$ is the Rydberg constant for hydrogen, $n$ is the principal quantum number corresponding to the initial energy level, and $\lambda$ is the wavelength of the spectral line.

Using this formula, the wavelength of the H$_\alpha$ line is:

$ \frac{1}{\lambda_{\alpha}} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) $

And the wavelength of the H$_\beta$ line is:

$ \frac{1}{\lambda_{\beta}} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) $

Therefore, the ratio of the wavelengths of the H$_\alpha$ and H$_\beta$ lines is:

$ \frac{\lambda_{\alpha}}{\lambda_{\beta}} = \frac{\left( \frac{1}{2^2} - \frac{1}{4^2} \right)}{\left( \frac{1}{2^2} - \frac{1}{3^2} \right)} = \frac{\frac{3}{16}}{\frac{5}{36}} = \frac{27}{20} $

So, comparing with the given ratio $\frac{x}{20}$, we find that $x = 27$.

The total binding energy of a nucleus is the binding energy per nucleon multiplied by the number of nucleons (protons and neutrons), which is the mass number.

The initial total binding energy of the nucleus is $242 \times 7.6 \, \text{MeV}$.

After the break, each fragment has a total binding energy of $121 \times 8.1 \, \text{MeV}$.

Since there are two such fragments, the final total binding energy is $2 \times 121 \times 8.1 \, \text{MeV}$.

The gain in binding energy is the final total binding energy minus the initial total binding energy. Therefore, the gain in binding energy is:

$2 \times 121 \times 8.1 \, \text{MeV} - 242 \times 7.6 \, \text{MeV} = 1960.2 \, \text{MeV} - 1839.2 \, \text{MeV} = 121 \, \text{MeV}$.

Therefore, the total gain in binding energy is $121 \, \text{MeV}$.

The binding energy of an electron in a hydrogen atom is given by the formula:

$ E = \frac{k e^2}{2 r} $

where:

• $E$ is the energy of the electron,
• $k$ is Coulomb's constant ($9 \times 10^9 \, \text{Nm}^2/\text{C}^2$),
• $e$ is the charge of the electron ($1.6 \times 10^{-19} \, \text{C}$), and
• $r$ is the radius of the orbit.

In this scenario, the energy $E$ required to separate a hydrogen atom into a proton and an electron is given as $12.8 \, \text{eV}$, which needs to be converted into joules using the conversion factor $1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$. So,

$ 12.8 \, \text{eV} = 12.8 \times 1.6 \times 10^{-19} \, \text{J} $

We can then substitute the given values into the energy equation and solve for $r$:

$ 12.8 \times 1.6 \times 10^{-19} \, \text{J} = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{2r} $

Solving for $r$, we get:

$ r = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{2 \times 12.8 \times 1.6 \times 10^{-19}} $

This simplifies to:

$ r = \frac{9 \times 10^{-10}}{16} $

Comparing this with the given form of the radius, which is $\frac{9}{x} \times 10^{-10}$, we find that the value of $x$ is 16.

The formula to calculate the radius of an orbit for a hydrogen-like atom/ion is:

$ r_n = r_0 \frac{n^2}{Z} $

where:

• $r_n$ is the radius of the nth orbit,
• $n$ is the principal quantum number (the orbit number),
• $r_0$ is the Bohr radius (radius of the first Bohr orbit in the hydrogen atom), and
• $Z$ is the atomic number (the number of protons in the nucleus).

We're dealing with a Li²⁺ ion and we're interested in the fifth orbit ($n = 5$), and given that $r_0$ is 0.51 Å and $Z$ for Li is 3, we can substitute these values into the formula:

$ r_5 = 0.51 \times \frac{25}{3} \text{ Å} = 4.25 \text{ Å} $

which is $4.25 \times 10^{-10}$ m, or equivalently $425 \times 10^{-12}$ m when converted to meters.

$ \Rightarrow {\lambda _{eff}} = {\lambda _1} + {\lambda _2}$

$ \Rightarrow {{\ln 2} \over {{t_{1/2}}}} = {{\ln 2} \over {{{({t_{1/2}})}_1}}} + {{\ln 2} \over {{{({t_{1/2}})}_2}}}$

$ \Rightarrow {t_{1/2}} = {{{{({t_{1/2}})}_1} \times {{({t_{1/2}})}_2}} \over {{{({t_{1/2}})}_1} + {{({t_{1/2}})}_2}}} = {{300 \times 30} \over {300 + 30}}s = {{300} \over {11}}s$

$ \Rightarrow \alpha = 300$

${T_{1/2}} = 150$ minutes

${A_0} = 64x$, where x is safe limit

$x = 64x \times {2^{ - {n \over {{T_{1/2}}}}}}$

$ \Rightarrow {1 \over {64}} = {2^{ - {n \over {{T_{1/2}}}}}}$

or ${n \over {{T_{1/2}}}} = 6$

$ \Rightarrow n = 6 \times 150$ minutes

= 15 hours

Energy released = Change in B.E.

(7.6 $\times$ 4) $-$ [4 $\times$ 1.1] = 26 MeV

${1 \over \lambda } = {R_H}\left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right)$

${1 \over {{\lambda _3}}} = {R_H}\left( {{1 \over {{3^2}}} - {1 \over {{6^2}}}} \right)$

${1 \over {{\lambda _2}}} = {R_H}\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right)$

$\therefore$ ${\lambda _3} - {\lambda _2} = a\lambda $

$a = 5$

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

${{{1 \over {{2^2}}} - {1 \over {{3^2}}}} \over {{1 \over {{2^2}}}}} = {x \over {x + 4}}$

$ \Rightarrow {{9 - 4} \over {9 \times 4 \times {1 \over 4}}} = {x \over {x + 4}} = {5 \over 9}$

$x = 5$

$N = {N_0}{e^{ - \lambda t}}$

$ \Rightarrow {{6.25} \over {100}} = {e^{ - \lambda t}}$

$ \Rightarrow {e^{ - \lambda t}} = {1 \over {16}} = {\left( {{1 \over 2}} \right)^4}$

$ \Rightarrow t = 4{t_{1/2}}$

$ \Rightarrow t = 20$ years

Impact parameter $\propto$ $\cot {\theta \over 2}$

$ \Rightarrow \sqrt {{{{d_1}} \over {{d_2}}}} = {{\sqrt 3 } \over 1}$

$ \Rightarrow {d_1} = 3{d_2}$

$ \Rightarrow x = 3$

E(in eV) = 13.6 $\times$ 9$\left( {1 - {1 \over 9}} \right)$

= 13.6 $\times$ 8 eV

$\Rightarrow$ $\lambda = {{12420} \over {13.6 \times 8}}\mathop A\limits^o $

= 114.15 $\mathop A\limits^o $