Atoms and Nuclei

Given $\rho=2.29 \times 10^{17} \mathrm{~kg} / \mathrm{m}^3$, mass $m=19.926 \times 10^{-27} \mathrm{~kg}$ $R_0=1.2 \times 10^{-15} \mathrm{~m}$ and $R=R_0 A^{\dfrac{1}{3}}$ Now use volume $=\dfrac{\text{ Mass }}{\text{ Density }}$ $\Rightarrow \quad \dfrac{4}{3} \pi R^3=\dfrac{M}{\rho}$ $\Rightarrow \quad \dfrac{4}{3} \pi\left[R_0(A)^{1 / 3}\right]^3=\dfrac{M}{\rho}$ $\Rightarrow \quad \dfrac{4}{3} \pi R_0^3 A=\dfrac{M}{\rho}$ $\Rightarrow \quad A=\dfrac{M}{\rho} \times \dfrac{3}{4 \pi \times R_0^3}=\dfrac{19.926 \times 10^{-27} \times 3}{2.29 \times 10^{17} \times 12.56 \times\left(1.2 \times 10^{-15}\right)^3} \approx 12$

When a particle of mass $m$ is subject to a constant force $F$ pulling it toward the origin, we can apply the Bohr model to describe its motion.

Starting with the relationship for centripetal force: $F = \dfrac{m v^2}{r}$ We can rearrange this to show: $\dfrac{v^2}{r} = \text{constant}$ From this, it follows that: $r \propto v^2 \quad \text{...(1)}$ In the Bohr model, the angular momentum $L$ is quantized and given by: $L = m v r = \dfrac{n h}{2 \pi} \quad \text{...(2)}$ Solving equations (1) and (2) together, we find: $v \propto n^{1/3}$ Substituting this into equation (1), we get: $r \propto n^{2/3}$ Thus, the radius of the $n^{\text{th}}$ orbit and the speed of the particle depend on $n$ as follows: $r \propto n^{2/3}$ $v \propto n^{1/3}$

The spectral series corresponds to groups of wavelengths that are produced when electrons in an atom make transitions between energy levels.

Each series is named after the scientist who discovered it and is characterized by the energy level to which the electrons transition.

The transitions to the level

define the Brackett series. Electrons that fall to the fourth energy level (from higher levels with

) emit or absorb radiation in the infrared region of the electromagnetic spectrum.

Here's a quick summary of the different spectral series: Pfund series: Transitions to

Brackett series: Transitions to

Lyman series: Transitions to

Balmer series: Transitions to

Therefore, the correct answer is: Option B: Brackett series

Water is used as a coolant in a nuclear reactor primarily because of its high specific heat capacity.

High specific heat capacity means that water can absorb a large amount of heat energy with only a small increase in its own temperature.

This property makes water an effective medium for transferring heat away from the reactor core, thereby maintaining safe operational temperatures and preventing overheating.

Therefore, the correct answer is: Option B: high specific heat capacity

Comparing (2) and (3)

Comparing (1) and (2) $$3 \lambda_1=2 \lambda_2 \quad \lambda_1

To determine the mass number and atomic number of the final product

in the provided nuclear reactions, we need to analyze the impact of each type of decay (alpha, positron emission, beta-minus decay, and electron capture) on the mass number and atomic number of the initial element

: 1. Alpha decay ($\alpha$ decay): In alpha decay, an alpha particle (which is a

nucleus) is emitted. This reduces the mass number by 4 units and the atomic number by 2 units.

2. Beta plus decay (positron emission,

): During positron emission, a proton in the nucleus is transformed into a neutron, and a positron is emitted.

This decreases the atomic number by 1 but does not change the mass number.

3. Beta-minus decay (

decay): In a beta-minus decay, a neutron in the nucleus converts into a proton and an electron (beta particle) and an antineutrino are emitted.

This results in an increase in the atomic number by 1, while the mass number remains unchanged.

4. Electron capture (

capture): During electron capture, an atomic electron is absorbed by the nucleus, causing a proton to convert into a neutron.

This process decreases the atomic number by 1 without altering the mass number.

From the calculations above, the mass number of

is 286, and its atomic number is 79.

Comparing these values with the multiple choices: Option A: Mass number = 280, Atomic number = 81 Option B: Mass number = 286, Atomic number = 80 Option C: Mass number = 288, Atomic number = 82 Option D: Mass number = 286, Atomic number = 81 None of these match exactly, implying an error in the problem or the answer options.

Based on the correct atomic number calculations and assuming option D is meant to be atomic number 79, then the corrected choice would have been: D: Mass number = 286, Atomic number = 79

The energy levels of a hydrogen atom are given by the formula :

where $E_n$ is the energy of the $n$-th energy level.

The ground state of hydrogen ($n=1$) has an energy of $-13.6 ~\mathrm{eV}$ as mentioned, which means that it would take $+13.6 ~\mathrm{eV}$ to ionize it (remove the electron completely) from this state, since ionization implies moving the electron to a state of zero energy.

The second excited state of hydrogen is when $n=3$ (as $n=1$ is the ground state and $n=2$ is the first excited state).

Thus, the energy of the second excited state is :

Since ionization implies moving the electron from its current energy level to $0$ energy, the energy required to ionize the atom from this state is the absolute value of its current energy state.

So, it will take $+1.51 ~\mathrm{eV}$ to ionize a hydrogen atom from its second excited state.

So, the correct answer is Option C : $1.51 ~\mathrm{eV}$.

Range of $\lambda$ is

to

which lies in U.V. region.

The given angular momentum of the electron is

, where $h$ is Planck's constant.

According to the Bohr model of the hydrogen atom, the allowed angular momenta for an electron are quantized and given by

where $m$ is the mass of the electron, $v$ is its velocity, $r$ is the radius of the orbit, and $n$ is the principal quantum number (an integer).

By comparing the given angular momentum with the quantized form, we get

, which simplifies to

. Now, we use the formula for the energy levels of the hydrogen atom:

. Since $n = 3$ , the energy can be calculated as

, which is approximately

.