Structure of Atom

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

where n is the principal quantum number (Bohr orbit number).

To find the energy in the third Bohr orbit (n = 3), we can use the formula:

Now, we are given the energy of the electron in the first Bohr orbit (n = 1):

First, let's convert this energy to electron volts (eV):

Now we can find the ratio of

to

:

Thus, the energy of the electron in the third Bohr orbit is

th of its energy in the first Bohr orbit.

Given are two statements : Assertion A : "In the photoelectric effect, the electrons are ejected from the metal surface as soon as the beam of light of frequency greater than threshold frequency strikes the surface." Reason R : "When the photon of any energy strikes an electron in the atom, transfer of energy from the photon to the electron takes place." From the understanding of the photoelectric effect, we know that the light beam must have a frequency $v$ that is greater than the threshold frequency $v_0$ for electrons to be ejected from the metal surface.

Therefore, Assertion A is correct.

However, Reason R is somewhat misleading in the context of the photoelectric effect.

It's true that a photon transfers its energy to an electron when it strikes the electron.

However, for the electron to be ejected (i.e., the photoelectric effect to occur), the energy of the photon, given by $E_{photon} = h \cdot v$, where $h$ is Planck's constant and $v$ is the frequency of the light, must be greater than the work function $\phi$ (the minimum energy needed to eject an electron) of the metal.

This condition is not captured in Reason R.

Therefore, the correct answer is : Option C : Assertion A is correct, but Reason R is not correct.

Option A: Nuclear charge $(\mathrm{z})$ is the correct answer.

Isoelectronic species are atoms and ions that have the same number of electrons.

When comparing the sizes of isoelectronic species, the major factor that affects their size is the nuclear charge $(\mathrm{z})$, which is the charge of the nucleus or the number of protons in the nucleus.

Here’s why: The more protons there are in the nucleus (higher nuclear charge), the more positively charged the nucleus is, and thus the stronger it can attract the electrons towards it.

This results in a smaller radius for the species.

Conversely, if there are fewer protons in the nucleus, the attraction to the electrons is weaker, leading to a larger radius.

Let's compare the species given: $\mathrm{F}^{-}$ (fluoride ion) has 9 protons in the nucleus and 10 electrons. $\mathrm{Ne}$ (neon atom) has 10 protons in the nucleus and 10 electrons. $\mathrm{Na}^{+}$ (sodium ion) has 11 protons in the nucleus and 10 electrons.

Despite having the same number of electrons, the $\mathrm{Na}^{+}$ ion will be the smallest among the three due to the higher nuclear charge of 11 protons, which attracts the same number of electrons more strongly than $\mathrm{Ne}$ with 10 protons and $\mathrm{F}^{-}$ with 9 protons.

Hence, $\mathrm{Na}^{+}$ has a lower atomic radius compared to $\mathrm{Ne}$, which in turn is smaller than $\mathrm{F}^{-}$.

Considering Option B (size being the same), Option C (electron-electron interaction), and Option D (principal quantum number), none of these factors are as significant as the difference in nuclear charge for isoelectronic species.

The principal quantum number (n) is the same for all these isoelectronic species since they all have their outer electrons in the second energy level (n=2) in the ground state.

Electron-electron interactions do contribute to energy levels and could affect the size slightly, but the primary and determining factor for isoelectronic species remains the nuclear charge, as it directly influences the effective nuclear charge felt by the electrons.

Therefore, Option A is the best answer.

The four quantum numbers for the outermost electron of potassium can be determined by first understanding the electronic configuration of potassium (atomic number 19).

The electron configuration is as follows:

The outermost electron resides in the 4s orbital.

The four quantum numbers are as follows: Principal quantum number ($n$): This indicates the energy level or shell number of the electron and for the 4s orbital, $n = 4$.

Azimuthal quantum number ($l$): This indicates the subshell or shape of the orbital.

For an $s$ orbital, $l = 0$.

Magnetic quantum number ($m$): This indicates the orientation of the orbital in space.

Since there is only one orientation for $s$ orbitals, $m = 0$.

Spin quantum number ($s$): This can be either $+\dfrac{1}{2}$ or $-\dfrac{1}{2}$.

Typically, the first electron in an orbital has a spin of $+\dfrac{1}{2}$.

Thus, the four quantum numbers for the outermost electron of potassium are:

This matches with Option B:

The correct set of four quantum numbers for the valence electron of a rubidium atom ($Z = 37$) is: Principal quantum number ($n$): This indicates the main energy level or shell of the electron.

For rubidium, the valence electron is in the 5th energy level, so $n = 5$.

Azimuthal quantum number ($l$): This defines the subshell or orbital type.

The possible values for $l$ range from 0 to $n-1$.

For rubidium, the valence electron is in the $5s$ orbital, so $l = 0$.

Magnetic quantum number ($m_l$): This indicates the orientation of the orbital in space.

For $l = 0$, the only possible value is $m_l = 0$.

Spin quantum number ($m_s$): This represents the spin of the electron, which can be either $+\dfrac{1}{2}$ or $-\dfrac{1}{2}$.

Generally, when an electron is added to an unoccupied orbital, it starts with a spin of $+\dfrac{1}{2}$.

Therefore, the correct set of quantum numbers for the valence electron of rubidium is: Option B:

A - II, B - IV, C - III, D - I Fact based.

Degenerate Orbitals: Orbitals that have the same energy are called degenerate orbitals.

Hydrogen-like Atoms (Single-Electron Systems): In a hydrogen atom (or any one-electron system), the energy of an orbital depends only on the principal quantum number $n$, not on the azimuthal quantum number $l$.

Therefore, all orbitals with the same $n$ (e.g., 3s, 3p, 3d) have the same energy and are degenerate.

Multi-electron Atoms: In atoms with more than one electron, the energy also depends on $l$ (and other factors like screening), so 3s, 3p, and 3d orbitals are not all degenerate in multi-electron atoms.

Now, let us evaluate the given statements: Statement (I): “The orbitals having same energy are called as degenerate orbitals.”

This statement is true by definition.

Statement (II): “In hydrogen atom, 3p and 3d orbitals are not degenerate orbitals.”

In a hydrogen atom, all orbitals with the same principal quantum number $n$ (like 3s, 3p, 3d) are degenerate.

Hence, this statement is false.

Therefore, the correct option is: Option A: Statement I is true but Statement II is false.

The postulate that contradicts quantum mechanics is Option A “The electron in a H atom’s stationary state moves in a circle around the nucleus.”

In quantum mechanics an electron in a stationary state is described by a delocalized wave‐function (an orbital), not by a definite circular orbit.

Matter consists of non-divisible atoms.

All the atoms of given element have same properties including mass.

Hence, statements (B) and (D) are incorrect.

To determine which spectral line of the hydrogen atom suits the wavelength for heat treatment of muscular pain (900 nm), we begin by applying the Rydberg equation: $\dfrac{1}{\lambda} = \mathrm{R}_{\mathrm{H}} \mathrm{Z}^2 \left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$ Given: $\lambda = 900 \text{ nm} = 9 \times 10^{-5} \text{ cm}$ $\mathrm{R}_{\mathrm{H}} = 10^5 \text{ cm}^{-1}$ Hydrogen atom (Z = 1) We proceed by solving the equation: $\dfrac{1}{\lambda \times \mathrm{R}_{\mathrm{H}}} = \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}$ Substituting the known values: $\dfrac{1}{9 \times 10^{-5} \text{ cm} \times 10^5 \text{ cm}^{-1}} = \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}$ This simplifies to: $\dfrac{1}{9} = \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}$ This condition is satisfied when $n_1 = 3$ and $n_2 = \infty$.

Therefore, the appropriate spectral transition is from $n_2 = \infty$ to $n_1 = 3$, which corresponds to the Paschen series transition $\infty \rightarrow 3$.