Atoms and Nuclei

To determine the ratio of the minimum wavelength in the Balmer series to the maximum wavelength in the Brackett series of the hydrogen spectrum, we start with the following calculations:

Maximum Wavelength of the Brackett Series:

The formula for the maximum wavelength in the Brackett series is given by:

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

Simplifying this, we have:

$ \frac{1}{\lambda_{\max}} = R \left[\frac{1}{16} - \frac{1}{25}\right] = \frac{9R}{25 \times 16} $

Minimum Wavelength of the Balmer Series:

For the Balmer series, the formula for the minimum wavelength is:

$ \frac{1}{\lambda_{\min}} = R \left(\frac{1}{2^2} - \frac{1}{\infty}\right) = \frac{R}{4} $

Ratio of Wavelengths:

We can now find the ratio of the minimum wavelength in the Balmer series to the maximum wavelength in the Brackett series:

$ \frac{\text{Minimum wavelength of Balmer series}}{\text{Maximum wavelength of Brackett series}} = \frac{9R}{25 \times 16} \times \frac{4}{R} = \frac{9}{100} $

Therefore, the ratio is:

$ 9:100 $

To find the age of the sample where the radioactive element $ A $ and the stable element $ B $ are present in a 1:15 ratio, we start by using the information given:

The ratio of $ A $ (denoted as $ N_X $) to $ B $ (denoted as $ N_Y $) is $ \frac{1}{15} $.

From this ratio:

$ \frac{N_X}{N_Y} = \frac{1}{15} $

The total presence of elements $ A $ and $ B $ can be written as:

$ N_X + N_Y = 16N_X $

We deduce that:

$ \frac{N_X}{N_X + N_Y} = \frac{1}{16} $

Rearranging, we have:

$ N_X = \frac{1}{16} (N_X + N_Y) $

Which implies:

$ N_X = \frac{1}{2^4} (N_X + N_Y) $

This result indicates that only $\frac{1}{16}$ of the original $ A $ remains. It is now a problem of calculating how many half-lives this corresponds to. Using the relationship that each half-life divides the remaining sample by 2:

After 1 half-life, the remaining amount is $ \frac{1}{2} $.

After 2 half-lives, the remaining amount is $ \frac{1}{4} $.

After 3 half-lives, the remaining amount is $ \frac{1}{8} $.

After 4 half-lives, the remaining amount is $ \frac{1}{16} $.

Thus, 4 half-lives have passed. With a half-life period of 62 years for element $ A $:

The age of the sample is:

$ 4 \times 62 = 248 \text{ years} $

Given:

Potential energy, $ \text{PE} = -6.8 \text{ eV} $

According to Bohr's postulate, the relationship between total energy $ E $ and kinetic energy $ K $ is:

$ E = K + \text{PE} $

We also know from Bohr's postulate:

$ E = \frac{\text{PE}}{2} $

Substituting the given potential energy:

$ E = \frac{-6.8 \text{ eV}}{2} = -3.4 \text{ eV} $

Using Bohr's relationship between kinetic and total energy:

$ -E = K $

Thus:

$ K = 3.4 \text{ eV} $

Given in equation form, where $ m_e $ is the mass of the electron and $ v_e $ is the speed of the electron:

$ K = \frac{1}{2} m_e v_e^2 $

Converting to joules:

$ 3.4 \times 1.6 \times 10^{-19} = \frac{1}{2} \times 9.1 \times 10^{-31} \times v_e^2 $

Solving for $ v_e $:

$ v_e = 1.093 \times 10^6 \text{ m/s} $

Expressing this speed in terms of the speed of light, $ c $:

$ v_e = \frac{c}{274} $

After calculating:

$ \frac{c}{274} = \frac{3 \times 10^8}{274} = 1.093 \times 10^6 \text{ m/s} $

Given:

The surface areas of two nuclei are in the ratio $9:25$.

This can be expressed mathematically as:

$ \frac{A_1}{A_2} = \frac{4\pi r_1^2}{4\pi r_2^2} = \frac{9}{25} $

Simplifying, we get:

$ \frac{r_1^2}{r_2^2} = \frac{9}{25} $

Taking the square root of both sides, we find the ratio of the radii:

$ \frac{r_1}{r_2} = \frac{3}{5} \tag{i} $

Assuming the density of the nuclei is $\rho$, the ratio of their masses can be calculated using their volumes:

$ \frac{m_1}{m_2} = \frac{V_1 \cdot \rho}{V_2 \cdot \rho} = \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3} $

Substituting the radius ratio:

$ \frac{m_1}{m_2} = \left(\frac{r_1}{r_2}\right)^3 = \left(\frac{3}{5}\right)^3 = \frac{27}{125} $

Thus, the mass number ratio of the nuclei is:

$ m_1 : m_2 = 27:125 $

Given that the ground state energy of a hydrogen atom is $-13.6 \text{ eV}$, this represents the total energy of the atom.

The kinetic energy (KE) of the electron in this state is equal to the negative of the total energy:

$ \mathrm{KE} = -E = -(-13.6) = 13.6 \text{ eV} $

The potential energy (PE) is calculated as the negative of two times the kinetic energy:

$ \begin{aligned} \mathrm{PE} & = -2 \times \mathrm{KE} \\ & = -2 \times 13.6 \\ & = -27.2 \text{ eV} \end{aligned} $

To determine the energy released in the fission of 0.1 kg of $\text{U}_{92}^{235}$, we start by calculating the number of atoms in this amount. The number of atoms in 235 g of uranium is $6.023 \times 10^{23}$.

Calculate the number of atoms in 100 g of uranium:

$ \text{Number of atoms} = \frac{6.023 \times 10^{23}}{235} \times 100 $

Given that the energy released per fission is 200 MeV, the total energy released is:

$ E_T = \frac{6.023 \times 10^{23} \times 200 \times 100}{235} \text{ MeV} $

Convert energy from MeV to Joules (since $1 \text{ MeV} = 1.6 \times 10^{-13} \text{ J}$):

$ E_T = \frac{6.023 \times 10^{23} \times 200 \times 1.6 \times 10^{-13} \times 100}{235} \text{ J} $

Convert energy from Joules to kilowatt-hours (since $1 \text{ kWh} = 3.6 \times 10^6 \text{ J}$):

$ E_T = \frac{6.023 \times 10^{23} \times 200 \times 1.6 \times 10^{-13} \times 100}{235 \times 3.6 \times 10^6} \text{ kWh} $

Simplify the calculation:

$ E_T = 0.02278 \times 10^6 \times 10^2 $

The final energy released in kilowatt-hours is:

$ E_T = 22.8 \times 10^5 \text{ kWh} $

Ultraviolet rays are produced when inner shell electrons in atoms move from one energy level to a lower level.

To calculate the speed of the electron in a hydrogen atom when it is in the $ n=3 $ energy level, we use the formula:

$ v_n = \frac{e^2}{2 \varepsilon_0 h n} $

where:

$ e = 1.6 \times 10^{-19} \mathrm{C} $ is the charge of the electron,

$ \varepsilon_0 = 8.85 \times 10^{-12} \mathrm{F/m} $ is the permittivity of free space,

$ h = 6.64 \times 10^{-34} \mathrm{Js} $ is Planck's constant, and

$ n = 3 $ is the principal quantum number for the energy level.

Substituting these values into the formula, we calculate:

$ v_3 = \frac{(1.6 \times 10^{-19})^2}{2 \times 8.85 \times 10^{-12} \times 6.64 \times 10^{-34} \times 3} $

Simplifying this, we find:

$ v_3 \approx 0.0073 \times 10^8 \mathrm{~m/s} $

Thus, the speed of the electron in the $ n=3 $ level is:

$ v_3 = 7.3 \times 10^5 \mathrm{~m/s} $

To determine the decay constant for one mole of radium, we start with its given activity:

Activity (A) equation:

$ A = \frac{1}{3.7} \times 10^3 \text{ curie} $

Key constants:

1 Curie = $ 3.7 \times 10^{10} $ decays per second

Avogadro's Number, $ N_A = 6 \times 10^{23} \text{ atoms/mol} $

Calculate Decay Constant ($\lambda$):

The decay constant $\lambda$ can be found using the formula:

$ \lambda = \frac{A}{N_A} $

Substitute the values:

$ A = \frac{1}{3.7} \times 10^3 \times 3.7 \times 10^{10} \text{ decays/s} $

Therefore:

$ \lambda = \frac{\frac{1}{3.7} \times 10^3 \times 3.7 \times 10^{10}}{6 \times 10^{23}} $

Solving this gives:

$ \lambda = \frac{1}{6} \times 10^{-10} \text{ s}^{-1} $

Hence, the decay constant of radium is calculated to be $\frac{1}{6} \times 10^{-10} \text{ s}^{-1}$.

The Bohr radius $ r_n $ for the $ n $-th orbit in a hydrogen atom is defined as

$ r_n = n^2 \frac{\hbar^2}{m_e e^2} $

For the first Bohr orbit ( $ n=1 $ ), the formula simplifies to:

$ r_1 = \frac{\hbar^2}{m_e e^2} $

Here, $ \hbar $ is the reduced Planck's constant.

Now, consider the Bohr radius for the $\mu$-meson system:

$ r_n^{\prime} = n^2 \frac{\hbar^2}{m_\mu Z e^2} $

Given that $ m_\mu=208 m_e $ and $ Z=3 $, we want to find the value of $ n $ such that the $\mu$-meson's orbit $ r_n^{\prime} $ is approximately equal to the radius of the first Bohr orbit $ r_1 $ of hydrogen:

$ \begin{align*} r_n^{\prime} &= r_1 \\ n^2 \frac{\hbar^2}{208 m_e 3 e^2} &= \frac{\hbar^2}{m_e e^2} \\ n^2 \frac{1}{208 \times 3} &= 1 \\ n^2 &= 208 \times 3 = 624 \\ n &= \sqrt{624} \approx 25 \end{align*} $

Hence, the quantum state $ n $ is approximately 25.

An alpha particle has an atomic mass number of 4. Let's assume the original nucleus has an atomic mass number $ A $. After losing 2 alpha particles, the new nucleus will have an atomic mass number $ A' = A - 2 \times 4 = A - 8 $.

The volume of a nucleus is proportional to its atomic mass number $ A $, i.e., $ V \propto A $.

Given that the volume of the new nucleus is 60 times that of an alpha particle, we write:

$ V_{\text{new}} = 60 \times V_{\text{alpha}} $

Thus,

$ A' = 60 \times 4 = 240 $

Since,

$ A' = A - 8 $

We equate:

$ A - 8 = 240 $

Solving for $ A $:

$ A = 240 + 8 = 248 $

Therefore, the atomic mass number $ A $ of the original nucleus is 248.

To determine the initial energy level $ n $ of a hydrogen atom that releases 12.75 eV when transitioning to the first energy level, follow these steps:

The change in energy ($ \Delta E $) for an electron falling from the $ n $-th level to the first level is given by:

$ \Delta E = -\frac{13.6}{n^2} - \left(-\frac{13.6}{1^2}\right) = 12.75 $

Simplify the equation:

$ -\frac{13.6}{n^2} + 13.6 = 12.75 $

Isolate the energy term:

$ -\frac{13.6}{n^2} = 12.75 - 13.6 $

Calculate the energy difference:

$ -\frac{13.6}{n^2} = -0.85 $

Rearrange to solve for $ n^2 $:

$ n^2 = \frac{13.6}{0.85} $

Calculate $ n^2 $:

$ n^2 = 16 $

Finally, find $ n $:

$ n = 4 $

Therefore, the $ n $-th orbit from which the electron falls is $ n = 4 $.

To determine the daily decrease in uranium mass in a reactor operating at a power of 12 MW, we start by calculating the total energy produced in one day.

Total Energy Per Day:

The power output of the reactor is 12 MW. Since power is energy per unit time, the energy output over one day is calculated as follows:

$ \text{Total energy per day} = 12 \times 10^6 \, \text{J/s} \times 24 \times 3600 \, \text{s} $

Energy Released Per Fission:

For each fission event of ${ }_{92}\mathrm{U}^{235}$, approximately 200 MeV of energy is released. To convert this energy to joules:

$ 200 \, \text{MeV} = 200 \times 1.6 \times 10^{-13} \, \text{J} = 3.2 \times 10^{-11} \, \text{J} $

Calculating Fissions Per Day:

Using the relationship between power, energy per fission, and the number of fissions, we get:

$ \text{Power} = \frac{\text{Number of Fissions} \times \text{Energy per Fission}}{\text{Time}} $

Solving for the number of fissions ($n$):

$ 12 \times 10^6 = \frac{n \times 3.2 \times 10^{-11}}{24 \times 3600} $

$ n = \frac{12 \times 10^6 \times 24 \times 3600}{3.2 \times 10^{-11}} = 3.2 \times 10^{22} $

Calculating Mass of Uranium Consumed:

To find how much uranium is consumed, we first determine the number of moles:

$ \text{Number of moles} = \frac{3.2 \times 10^{22}}{6.023 \times 10^{23}} = 0.053 $

The mass of uranium consumed is then:

$ \text{Mass of uranium} = 0.0531 \times 235 \approx 12.64 \, \text{g} $

Thus, the daily decrease in uranium mass in the reactor is approximately 12.64 grams.

Given:

The binding energy for a hydrogen atom is 13.6 eV.

We know:

The energy of an electron in the $ n $-th state is given by:

$ E_n = -13.6 \frac{Z^2}{n^2} $

For $\mathrm{Li}^{2+}$, the atomic number $ Z = 3 $.

To find the energy required to remove the electron from the first excited state ($ n = 2 $):

Substitute into the equation:

$ \begin{aligned} E_2 &= -13.6 \times \frac{3^2}{2^2} = -13.6 \times \frac{9}{4} \\ E_2 &= -30.6 \, \text{eV} \end{aligned} $

Since $ E_2 + E = 0 $, we have:

$ -30.6 + E = 0 $

Thus, solving for $ E $:

$ E = +30.6 \, \text{eV} $

Therefore, the energy required to remove the electron from the first excited state of $\mathrm{Li}^{2+}$ is 30.6 eV.

Given:

Initial amount of $ A_1 = 40 \, \text{g} $

Initial amount of $ A_2 = 160 \, \text{g} $

Half-life of $ A_1, T_{A_1} = 20 \, \text{s} $

Half-life of $ A_2, T_{A_2} = 10 \, \text{s} $

To determine when the amounts of $ A_1 $ and $ A_2 $ will be equal, we use the decay formula:

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

For $ A_1 $, the remaining amount is:

$ N_{A_1} = 40 \left(\frac{1}{2}\right)^{t/20} $

For $ A_2 $, the remaining amount is:

$ N_{A_2} = 160 \left(\frac{1}{2}\right)^{t/10} $

To find the time when these amounts are equal, set $ N_{A_1} = N_{A_2} $:

$ \begin{aligned} 40 \left(\frac{1}{2}\right)^{t/20} &= 160 \left(\frac{1}{2}\right)^{t/10} \\ 2^{t/10} &= 4 \cdot 2^{t/20} \Rightarrow 2^{\frac{t}{10}-\frac{t}{20}} = 4 \\ 2^{t/20} &= 2^2 \Rightarrow \frac{t}{20} = 2 \\ t &= 40 \, \text{s} \end{aligned} $

Therefore, the amounts of $ A_1 $ and $ A_2 $ in the mixture will become equal after 40 seconds.

To determine which transitions of a $\mathrm{He}^{+}$ ion can produce a spectral line with the same wavelength as a given transition in a hydrogen atom, let's examine the Rydberg formula:

For a hydrogen atom ($\mathrm{Z}=1$), the wavelength $\lambda_1$ for a transition from $n_2$ to $n_1$ is given by:

$ \frac{1}{\lambda_1} = R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) $

Here, $R$ is the Rydberg constant, and $n_1$ and $n_2$ are the principal quantum numbers with $n_2 > n_1$.

For a $\mathrm{He}^{+}$ ion (with $Z=2$), if the electron transitions from $n_4$ to $n_3$, the wavelength $\lambda_2$ is given by:

$ \frac{1}{\lambda_2} = R \cdot 4 \left(\frac{1}{n_3^2} - \frac{1}{n_4^2}\right) $

Assuming $\lambda_1 = \lambda_2$, we have:

$ R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) = R \cdot 4 \left(\frac{1}{n_3^2} - \frac{1}{n_4^2}\right) $

By comparing both sides, we get the relationships:

$ n_1 = \frac{n_3}{2}, \quad n_2 = \frac{n_4}{2} $

Both $n_1$, $n_2$, $n_3$, and $n_4$ must be integers, as they represent principal quantum numbers.

This condition is satisfied when $n_3 = 2$ and $n_4 = 4$, ensuring that transitions corresponding to these quantum numbers yield the same wavelength for both $\mathrm{He}^{+}$ and hydrogen.

We are given that the half-life of A is the same as the average life of B. The relationship between half-life ($T_{1/2}$) and the decay constant ($\lambda$) is:

$T_{1/2} = \frac{\ln 2}{\lambda}$

For the average life ($\tau$), the relationship with the decay constant is:

$\tau = \frac{1}{\lambda}$

According to the given information, the half-life of A is equal to the average life of B:

$T_{1/2(A)} = \tau_{B}$

Now, we can substitute the relationships for half-life and average life:

$\frac{\ln 2}{\lambda_{A}} = \frac{1}{\lambda_{B}}$

To find the correct relationship between $\lambda_{A}$ and $\lambda_{B}$, we can rearrange the equation:

$\lambda_{A} = \lambda_{B} \ln 2$

Let's use the formula for the remaining mass of a radioactive substance after a certain time:

$N(t) = N_0(1/2)^{t/T}$

where N(t) is the mass at time t, N₀ is the initial mass, T is the half-life, and t is the time elapsed.

We are given that $\frac{7}{8}$th of the original mass has disintegrated. Therefore, the remaining mass is $\frac{1}{8}$th of the original mass:

$\frac{N(t)}{N_0} = \frac{1}{8}$

Using the formula, we have:

$\frac{1}{8} = (1/2)^{t/T}$

Taking the logarithm of both sides:

$\log_{1/2}\frac{1}{8} = \frac{t}{T}$

$3 = \frac{t}{T}$

Now, solving for t:

$t = 3T$

So, the time taken for disintegrating $\frac{7}{8}$th part of the original mass is 3T.

According to Bohr's model of the hydrogen atom, the angular momentum of an electron in an orbit is an integral multiple of Planck's constant divided by $2\pi$ (or $h/2\pi$, where $h$ is the Planck's constant). This can be expressed as:

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

where $n$ is the principal quantum number or the orbit number.

So, for the first orbit ($n=1$), the angular momentum $L_1$ is:

$ L_1 = 1 \times \frac{h}{2\pi} = \frac{h}{2\pi} $

And for the second orbit ($n=2$), the angular momentum $L_2$ is:

$ L_2 = 2 \times \frac{h}{2\pi} = \frac{h}{\pi} $

The change in angular momentum when moving from the first to the second orbit is the difference between $L_2$ and $L_1$:

$ \Delta L = L_2 - L_1 = \frac{h}{\pi} - \frac{h}{2\pi} = \frac{h}{2\pi} $

Since $\frac{h}{2\pi}$ is equal to the initial angular momentum $L_1$, the change in angular momentum when the electron moves to the second orbit is $L$.

Therefore, the correct answer is $L$.

The decay of a radioactive material follows an exponential decay law, which can be expressed as:

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

where:

• $N$ is the final amount of the material,
• $N_0$ is the initial amount of the material,
• $t$ is the elapsed time,
• $T$ is the half-life of the material.

From the problem, we know that the material is reduced to $1/8$ of its original amount in 3 days. Therefore, the half-life of the material can be calculated as follows:

$ \frac{1}{8} = \left(\frac{1}{2}\right)^{\frac{3}{T}} $

This simplifies to $2^{-3} = 2^{-\frac{3}{T}} $, which gives $ T = 1 \, \text{day} $.

Knowing the half-life, we can now find the initial amount of the material. We know that after 5 days, $8 \times 10^{-3} \, \text{kg}$ of the material is left. Therefore, we can write:

$ 8 \times 10^{-3} \, \text{kg} = N_0 \cdot \left(\frac{1}{2}\right)^{\frac{5}{1}} $

Solving this equation for $N_0$ gives:

$ N_0 = 8 \times 10^{-3} \, \text{kg} \cdot 2^{5} = 256 \times 10^{-3} \, \text{kg} = 0.256 \, \text{kg} = 256 \, \text{g} $

When a metal target is bombarded with high-energy electrons, the phenomenon known as X-ray emission occurs. This is due to the excitation of the inner-shell electrons in the metal atoms by the high-energy electrons. When these inner-shell electrons drop back to their original energy levels, they emit energy in the form of X-ray photons.

In the semi-empirical mass formula, the terms have the following forms:

• The volume term (binding energy) is proportional to the volume of the nucleus, and therefore to the number of nucleons $A$.


• The surface term takes into account that the nucleons at the surface of the nucleus have fewer nearest neighbors than those in the interior, and so the binding energy is less for surface nucleons. This term is usually proportional to $A^{2/3}$.


• The Coulomb term arises because the protons in the nucleus repel each other. This term is usually proportional to $\frac{Z(Z-1)}{A^{1/3}}$.


• The symmetry term arises because of the Pauli exclusion principle, but it's not mentioned here.
• The pairing term arises because of the tendency of protons and neutrons to pair up, but it's not mentioned here.

Now, let's check the statements:

• A: The surface energy per nucleon $b_{s}=-a_{1} A^{2 / 3}$ is incorrect. The correct form should be proportional to $A^{2/3}$, but it should not be per nucleon.


• B: The Coulomb contribution to the binding energy $b_{c}=-a_{2} \frac{Z(Z-1)}{A^{4 / 3}}$ is incorrect. The correct form should be proportional to $\frac{Z(Z-1)}{A^{1/3}}$.


• C: The volume energy $b_{v}=a_{3} A$ is correct.


• D: Decrease in the binding energy is proportional to surface area is correct. This represents the surface term in the semi-empirical mass formula.


• E: While estimating the surface energy, it is assumed that each nucleon interacts with 12 nucleons is incorrect. While the exact number can vary depending on the model used, it's generally agreed that each nucleon interacts with a few nearest neighbors, not necessarily 12.

So, the correct answer is C, D only.

According to Bohr's quantization of angular momentum, the angular momentum $L$ of a particle in a circular orbit is given by:

$L = n\hbar$

Where $n$ is an integer and $\hbar$ is the reduced Planck's constant. The angular momentum $L$ can also be expressed as:

$L = mvr$

Where $m$ is the mass of the particle, $v$ is its linear velocity, and $r$ is the radius of the orbit.

Now, we are given the potential energy $U = \frac{1}{2} m\omega^2r^2$. Since the particle is in a circular orbit, its centripetal force is provided by the gradient of the potential energy:

$m\frac{v^2}{r} = -\frac{\mathrm{d}U}{\mathrm{d}r} = -m\omega^2r$

We can simplify this equation to get the relation between $v$ and $r$:

$v^2 = \omega^2r^2$

Now, let's combine the equations for angular momentum and the relation between $v$ and $r$:

$n\hbar = mvr = m\sqrt{\omega^2r^2}r = m\omega r^2$

We can now solve for the radius $r$ in terms of $n$:

$r^2 = \frac{n\hbar}{m\omega}$

Taking the square root of both sides, we get:

$r \propto \sqrt{n}$

So, the radius of the $n^{\text{th}}$ orbit is proportional to $\sqrt{n}$.

$R = {R_0}{A^{{1 \over 3}}}$, using this

$\rho = {M \over {{4 \over 3}\pi {R^3}}} = {{A{m_P}} \over {{4 \over 3}\pi R_0^3A}} = {{{m_P}} \over {{4 \over 3}\pi R_0^3}}$

$\rho$ is independent of mass number.

$\therefore$ A is false

$v\,\alpha {z \over n}$

${{{v_1}} \over {{v_2}}} = \left( {{{{n_2}} \over {{n_1}}}} \right)$

$ \Rightarrow {{3.6 \times {{10}^6}} \over {{v_2}}} = {3 \over 7}$

$ \Rightarrow {v_2} = {7 \over 3} \times 3.6 \times {10^6}$ m/s

$ = 8.4 \times {10^6}$ m/s

${n_A} = 20$ moles

${n_B} = 10$ moles

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

${N_A} = (20\,N){e^{ - \left( {{{\ln 2} \over 1} \times 2} \right)}}$

$ = {{20\,N} \over 4} = 5\,N$ (N = Avogadro's Number)

${N_B} = 10\,N\,{e^{ - 4\ln 2}}$

$ = \left( {{{10\,N} \over {16}}} \right)$

${N_A} + {N_B} = 5\,N + {{10\,N} \over {16}} = \left( {{{90\,N} \over {16}}} \right) = 3.38 \times {10^{24}}$