Structure of Atom

1. Principal Quantum Number ($n$): This number tells us the principal energy shell. In the notation of an orbital (like 3d or 4s), the number at the beginning is the value of $n$.

2. Azimuthal Quantum Number ($l$): This number defines the subshell and the shape of the orbital. We use specific letters to denote the value of $l$:

   • If $l = 0$, the subshell is 's'.

   • If $l = 1$, the subshell is 'p'.

   • If $l = 2$, the subshell is 'd'.

   • If $l = 3$, the subshell is 'f'.

The designation for an orbital is given by writing the value of $n$ followed by the letter for the subshell (determined by $l$).

Now, let's look at each item from List I.

A. For $n=2$ and $l=1$:

• The principal quantum number $n$ is 2.

• The azimuthal quantum number $l$ is 1, which corresponds to the 'p' subshell.

• Therefore, the orbital is designated as 2p.

• This matches with II in List II.

B. For $n=4$ and $l=0$:

• The principal quantum number $n$ is 4.

• The azimuthal quantum number $l$ is 0, which corresponds to the 's' subshell.

• Therefore, the orbital is designated as 4s.

• This matches with III in List II.

C. For $n=5$ and $l=3$:

• The principal quantum number $n$ is 5.

• The azimuthal quantum number $l$ is 3, which corresponds to the 'f' subshell.

• Therefore, the orbital is designated as 5f.

• This matches with IV in List II.

D. For $n=3$ and $l=2$:

• The principal quantum number $n$ is 3.

• The azimuthal quantum number $l$ is 2, which corresponds to the 'd' subshell.

• Therefore, the orbital is designated as 3d.

• This matches with I in List II.

So, the correct matches are:

• A → II

• B → III

• C → IV

• D → I

Looking at the options, we can see that this combination corresponds to Option C.

The correct answer is Option C: A-II, B-III, C-IV, D-I.

Energy = Power × Time

$ = 150~\mathrm{watt} \times 1~\mathrm{s} $

$ = 150~\mathrm{J} $

Only $ 8\% $ of this energy is converted into light energy.

$ \text{Energy as light} = \frac{150 \times 8}{100} = 12~\mathrm{J} $

We know that total light energy is equal to the number of photons multiplied by the energy of one photon.

$ E = n \times \text{Energy of one photon} $

Energy of one photon = $ 4.42 \times 10^{-19}~\mathrm{J} $

$ n = \frac{E}{\text{Energy of one photon}} = \frac{12}{4.42 \times 10^{-19}} $

$ n = 2.715 \times 10^{19} $

Hence, the bulb emits approximately $ 2.7 \times 10^{19} $ photons per second.

To find the ratio of the wavelengths of the light absorbed by a hydrogen atom during the transitions $ n=2 \rightarrow n=3 $ and $ n=4 \rightarrow n=6 $, we start by calculating the change in energy ($\Delta E$) for each transition.

The energy of an electron in a hydrogen atom at a given level $n$ is given by:

$ E_n = \frac{-R_H}{n^2} $

where $ R_H $ is the Rydberg constant.

Transition $ n=2 \rightarrow n=3 $:

$ \Delta E_{2 \rightarrow 3} = E_3 - E_2 = \frac{-R_H}{3^2} - \left(\frac{-R_H}{2^2}\right) $

$ = R_H\left(\frac{1}{4} - \frac{1}{9}\right) $

$ = R_H \times \frac{5}{36} $

The wavelength $\lambda_{2 \rightarrow 3}$ is then:

$ \lambda_{2 \rightarrow 3} = \frac{h c}{\Delta E_{2 \rightarrow 3}} = \frac{h c \cdot 36}{R_H \cdot 5} $

Transition $ n=4 \rightarrow n=6 $:

$ \Delta E_{4 \rightarrow 6} = E_6 - E_4 = \frac{-R_H}{36} + \frac{R_H}{16} $

$ = \frac{R_H \times 20}{36 \times 16} $

The wavelength $\lambda_{4 \rightarrow 6}$ is then:

$ \lambda_{4 \rightarrow 6} = \frac{h c}{\Delta E_{4 \rightarrow 6}} = \frac{h c \times 36 \times 16}{R_H \cdot 20} $

Calculating the ratio:

The ratio of the wavelengths $\frac{\lambda_{2 \rightarrow 3}}{\lambda_{4 \rightarrow 6}}$ is:

$ \frac{\lambda_{2 \rightarrow 3}}{\lambda_{4 \rightarrow 6}} = \frac{\frac{h c \cdot 36}{R_H \cdot 5}}{\frac{h c \times 36 \times 16}{R_H \cdot 20}} $

$ = \frac{1}{4} $

Therefore, the ratio of the wavelengths for the given transitions is $\frac{1}{4}$.

The energy and radius of the first Bohr orbit for a hydrogen-like atom is given by:

Energy:

$ E_n = \frac{-2.18 \times 10^{-18} \times Z^2}{n^2} \, \text{J} $

Radius:

$ r_n = \frac{52.9 \times n^2}{Z} \, \text{pm} $

Where:

$ Z $ is the atomic number.

$ n $ is the orbit number (or principal quantum number).

Let's calculate these values for the ions $\text{He}^+$ and $\text{Li}^{2+}$:

For $\text{He}^+$:

The atomic number, $ Z = 2 $

Principal quantum number, $ n = 1 $

Energy, $ E_{\text{He}^+} $:

$ E_{\text{He}^+} = -2.18 \times 10^{-18} \times 2^2 = -8.72 \times 10^{-18} \, \text{J} $

Radius, $ r_{\text{He}^+} $:

$ r_{\text{He}^+} = \frac{52.9 \times 1^2}{2} = 26.45 \, \text{pm} $

For $\text{Li}^{2+}$:

The atomic number, $ Z = 3 $

Principal quantum number, $ n = 1 $

Energy, $ E_{\text{Li}^{2+}} $:

$ E_{\text{Li}^{2+}} = -2.18 \times 10^{-18} \times 3^2 = -19.62 \times 10^{-18} \, \text{J} $

Radius, $ r_{\text{Li}^{2+}} $:

$ r_{\text{Li}^{2+}} = \frac{52.9 \times 1^2}{3} = 17.63 \, \text{pm} $

These calculations show the energy and radius for the first Bohr orbit of $\text{He}^+$ and $\text{Li}^{2+}$.

(I) $n=4, I=2, m I=-2, s=-\frac{1}{2}$; represents $4 d(n+I=6)$

(II) $n=3, I=2, m_I=1, s=+\frac{1}{2}$; represents $3 d(n+I=5)$

(III) $n=4, I=1, m_I=0, s=+\frac{1}{2}$; represents $4 p(n+I=5)$

(IV) $n=3, I=1, m_1=-1, s=+\frac{1}{2}$; represents $3 p(n+I=4)$

Order of energy depends on the $(n+I)$, greater is the $(n+I)$ value greater is the energy, if $(n+I)$ is same, then it depends on $n$; if '$n$' is more, energy is more.

Step-1: According to $(\mathrm{n}+\mathrm{I})$

$\text { Energy }=(\mathrm{I})>(\mathrm{II})=(\mathrm{III})>(\text { IV) }$

Step-2 : If $n \uparrow$, then energy increases

$\text { Energy }=\text { (I) }>\text { (III) }>\text { (II) }>\text { (IV) }$

For Balmer series

$\mathrm{n}_1=2$

$\mathrm{n}_2=3$

$\begin{aligned} & \bar{v}=\frac{1}{\lambda}=R_H\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right] \\ & \bar{v}=R_H\left[\frac{1}{4}-\frac{1}{9}\right] \\ & \bar{v}=\frac{5 R_H}{36} \mathrm{~cm}^{-1} \end{aligned}$

The energy levels of an electron in a hydrogen-like ion (an atom or ion with only one electron) can be quantified using the formula:

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

where:

• $ E_n $ is the energy of the electron in the nth energy level,


• $ Z $ is the atomic number (number of protons) of the ion,


• $ 13.6 \text{ eV} $ is the ionization energy of hydrogen,


• $ n $ is the principal quantum number (the energy level).

Since we need to compare this across different ions in different energy states, let's plug in some numbers:

For $ \mathrm{He}^{+} $ (Helium ion):

• $ Z = 2 $ (as helium has 2 protons)


• $ n = 1 $ for ground state

$ E_1 = -\frac{(2)^2 \cdot 13.6 \text{ eV}}{1^2} = -4 \cdot 13.6 \text{ eV} = -54.4 \text{ eV} $

However, the problem gives the energy in joules, and it's given a constant $ x $. So, $ x = 54.4 \text{ eV} $ (converted to joules as needed).

Next, for the $ \mathrm{Be}^{3+} $ ion:

• $ Z = 4 $ (as beryllium has 4 protons)


• $ n = 2 $

$ E_2 = -\frac{(4)^2 \cdot 13.6 \text{ eV}}{2^2} = -\frac{16 \cdot 13.6 \text{ eV}}{4} = -54.4 \text{ eV} $

Since initially $ x = 54.4 \text{ eV} $ (or its equivalent in joules) for $ \mathrm{He}^{+} \text{ at } n=1 $, and now we have the same energy for $ \mathrm{Be}^{3+} \text{ at } n=2 $, the energy level in joules would also equate to $ x $, just at a different energy state and ion.

So, the answer is:

Option A: $ -x $.

To match List I (Quantum Numbers) with List II (Information provided), we need to understand what each quantum number represents:

1. Principal Quantum Number ($n$): This quantum number determines the size and energy level of the orbital. An increase in $n$ implies a higher energy level and a larger orbital size.

2. Azimuthal Quantum Number ($\ell$): (not directly listed but related to $\mathrm{m_l}$ because $\mathrm{m_l}$ depends on $\ell$): This quantum number defines the shape of the orbital. Different values of $\ell$ correspond to different shapes (s, p, d, f, etc.).

3. Magnetic Quantum Number ($\mathrm{m_l}$): This quantum number describes the orientation of the orbital in space. It can take values from $-\ell$ to $\ell$, where each value corresponds to a specific orientation of the orbital.

4. Spin Quantum Number ($\mathrm{m_s}$): This quantum number specifies the orientation of the spin of the electron. The two possible values are $+\frac{1}{2}$ and $-\frac{1}{2}$, representing the two possible spin orientations (up or down).

Matching the given options:

A. $\mathrm{m_l}$ - should be matched with "Orientation of orbital" (III).

B. $\mathrm{m_s}$ - should be matched with "Orientation of spin of electron" (IV).

C. I - This is likely intended to be $\ell$, and would be matched with "Shape of orbital" (I).

D. $n$ - relates to "Size of orbital" (II), as it affects the distance from the nucleus along with the energy level.

Using the understanding above, we can identify the correct matches:

A-III (Orientation of orbital)

B-IV (Orientation of spin of electron)

C-I (Shape of orbital)

D-II (Size of orbital)

Therefore, the correct answer is:

Option B: A-III, B-IV, C-I, D-II

$ \mathrm{n}=5, \ell=2, \mathrm{~m}=-2,-1,+1,+2, \mathrm{~m}_{\mathrm{s}}=+\frac{1}{2} $

Statement I is correct because in quantum mechanics, the wave function $\psi$ indeed depends on the coordinates (position variables) of the electron in an atom. These coordinates describe where the electron is likely to be at any given time and the wave function varies with these coordinates.

Statement II, however, contains an inaccuracy. The probability of finding an electron at a point within an atom is not directly proportional to the orbital wave function $\psi$ itself, but rather to the square of its absolute value, $|\psi|^2$. This squared value is referred to as the probability density, and it is what provides the likelihood of locating an electron at a particular point in space.

It is statement based question.

Statements B, C & E are correct.

(B) Mass of the electron is $9.10939 \times 10^{-31} \mathrm{~kg}$

(C) All the isotopes of given elements show same chemical properties.

(E) Dalton's atomic theory, regarded the atom as an ultimate particle of matter.

Sol. Number of permissible values of magnetic quantum number for a given value of azimuthal quantum $(l)$

$\Rightarrow \mathrm{n}_{\mathrm{m}}=2 \ell+1$

$\Rightarrow \ell=\frac{\mathrm{n}_{m}-1}{2}$

Energy of one photon $ = {{hc} \over \lambda }$ ($\lambda$ = 300 nm)

For one mole photons, $E = {{hc} \over \lambda } \times {N_A}$

$E = {{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8} \times 6.023 \times {{10}^{23}}} \over {300 \times {{10}^{ - 9}}}}$

$E = 3.99 \times {10^5}$ J mol^$-$1

Kinetic energy $ = 1.68 \times {10^5}$ J mol^$-$1

${W_0} = E - K.E.$

$ = 3.99 \times {10^5} - 1.68 \times {10^5}$

$ = 2.31 \times {10^5}$ J mol^$-$1

$\bullet$ In an atom, all the five 3d orbitals are equal in energy in free state i.e., degenerate.

$\bullet$ The shape of d_{x² $-$ y²} is different then shape of d_z²

$\bullet$ The size of orbital depends on principal quantum number 'n' therefore all the five 3d orbitals are different in size when compared to the respective 4d orbitals.

$\bullet$ Shape of orbitals depends on azimuthal quantum number 'I' therefore shapes of 4d orbitals are similar to the respective 3d orbitals.

${r_n} \propto {{{n^2}} \over Z}$

${{{r_3}(L{i^{2 + }})} \over {{r_2}(H{e^ + })}} = {{{{({n_3})}^2}} \over {Z(L{i^{2 + }})}} \times {{Z(H{e^ + })} \over {{{({n_2})}^2}}}$

${{{r_3}(L{i^{2 + }})} \over {105.8}} = {{{{(3)}^2}} \over 3} \times {2 \over {{{(2)}^2}}}$

$ = 105.8 \times {3 \over 2}$

${r_3}(L{i^{2 + }}) = 158.7$ pm

For $n = 3$ and $l = 1$:

• $n = 3$ means the electron is in the third shell

• $l = 1$ means it is a $p$-subshell

Now, for a $p$-subshell:

• the possible values of $m_l$ are $-1, 0, +1$

• so, number of orbitals $= 3$

Each orbital can hold a maximum of $2$ electrons.

So total electrons in the $3p$ subshell:

$ 3 \times 2 = 6 $

Therefore, the correct answer is Option B: 6.