Atomic Structure

$ \begin{array}{lcl} \hline \text { Element } & \begin{array}{c} \text { Atomic } \\ \text { number } \end{array} & \begin{array}{l} \text { Electronic } \\ \text { configuration } \end{array} \\ \hline \mathrm{Fe} & 26 & 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^2 3 d^6 \\ \hline \mathrm{Cl} & 17 & 1 s^2 2 s^2 2 p^6 3 s^2 3 p^5 \\ \hline \mathrm{Mg} & 12 & 1 s^2 2 s^2 2 p^6 3 s^2 \\ \hline \mathrm{Ne} & 10 & 1 s^2 2 s^2 2 p^6 \\ \hline \text { Total number of } d \text { electrons in } \mathrm{Fe}=6 \\ \text { Total number of } s \text { electrons in } \mathrm{Mg}=6 \\ \text { Total number of } p \text { electrons in } \mathrm{Cl}=11 \\ \text { Total number of } p \text { electrons in } \mathrm{Ne}=6 \end{array} $

According to the Heisenberg Uncertainty Principle, the product of the uncertainties in position ($\Delta x$) and momentum ($\Delta p$) of a particle is given by:

$ \Delta x \cdot \Delta p = \frac{h}{4\pi} \quad \ldots \text{(i)} $

In this scenario, it is stated that the uncertainty in the position ($\Delta x$) is equal to the uncertainty in momentum ($\Delta p$). Therefore, we can express this as:

$ \Delta x = \Delta p \quad \text{or} \quad \Delta x = m \Delta v \quad \ldots \text{(ii)} $

where $m$ is the mass of the electron and $\Delta v$ is the uncertainty in velocity. Substituting this relationship into Eq. (i), we have:

$ (m \Delta v)(m \Delta v) = \frac{h}{4\pi} $

Simplifying this equation, we find:

$ (\Delta v)^2 = \frac{h}{4\pi m^2} $

Taking the square root of both sides, the uncertainty in velocity ($\Delta v$) is calculated as:

$ \Delta v = \frac{1}{2m} \sqrt{\frac{h}{\pi}} $

From the graph, $\left[\psi^2(r)\right.$ vs $\left.r\right]$

It touches $r$ axis at one point, which , means it has only one radical node and at $r=0$ it has some value which shows that it should be for ' $s$ ' orbital.

$\therefore n-l-1=1$ (where, $l=0$ )

$ \begin{aligned} n-1 & =1 \\ n & =2 \end{aligned} $

Hence, $2 s$-orbital.

For an electron in a stationary state of $\mathrm{Li}^{2+}(Z=3)$, we are given that the angular momentum is $\frac{3 h}{\pi}$.

Determining $n$:

Using Bohr's postulate for angular momentum:

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

Set this equal to the given angular momentum:

$ \frac{3 h}{\pi} = \frac{n h}{2 \pi} $

Solving for $n$:

$ 3 \times 2 = n \quad \Rightarrow \quad n = 6 $

Calculating the Radius:

The formula for the radius in Bohr's model is:

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

where $r_0 = 0.529 \text{ Å}$. Substituting the known values:

$ r = 0.529 \times \frac{36}{3} = 6.348 \text{ Å} $

Calculating the Energy:

The energy of the electron is given by:

$ E = -13.6 \text{ eV} \times \frac{Z^2}{n^2} $

Converting to joules:

$ E = -13.6 \times 1.6 \times 10^{-19} \times \frac{9}{36} $

$ E = -5.45 \times 10^{-19} \text{ J} $

Thus, the radius is $6.348 \text{ Å}$ and the energy is $-5.45 \times 10^{-19} \text{ J}$.

Let's analyze the electron configurations of the given elements, focusing on the electrons in the (n – 1) shell. For transition elements in period 4, the outermost electrons are in the 4s orbital (n = 4), and the immediately inner electrons in the 3d orbitals belong to the (n – 1) = 3 shell.

For transition metals, we interpret “(n – 1) shell” as referring to the electrons in the 3d orbital (since n = 4 for period 4 elements).

Look at the electron configurations:

$\textbf{Mn (atomic no. 25):} \quad [Ar]\,3d^5\,4s^2$

$\textbf{Cr (atomic no. 24):} \quad [Ar]\,3d^5\,4s^1$

Even though Cr shows an exception (its 4s electron count is 1 instead of 2 due to extra stability of a half-filled 3d subshell), what matters here is the number of electrons in the 3d orbital.

Both Mn and Cr have:

$\textbf{3d electrons: }3d^5$

Checking the other options:

$\textbf{Fe: } [Ar]\,3d^6\,4s^2$ and $\textbf{Mn: } [Ar]\,3d^5\,4s^2$ have different numbers of 3d electrons (6 vs. 5).

$\textbf{Zn: } [Ar]\,3d^{10}\,4s^2$ and $\textbf{Fe: } [Ar]\,3d^6\,4s^2$ (10 vs. 6 in 3d).

$\textbf{K (atomic no. 19):} \quad [Ar]\,4s^1$ has no 3d electrons, while $\textbf{Sc (atomic no. 21):}\quad [Ar]\,3d^1\,4s^2$ has one.

Thus, the only pair with the same number of electrons in the (n – 1) (i.e., the 3d) shell is Mn and Cr.

Therefore, the answer is:

Option D (Mn, Cr).

To determine the wavelength of the emitted electron, we can use the following process:

First, calculate the energy absorbed by the electron:

$ J = 1.5 \times 2.18 \times 10^{-18} \, \text{J} = 3.27 \times 10^{-18} \, \text{J} \, \text{(i)} $

Using Einstein's photoelectric equation, we find the excess energy, which is the energy used by the emitted electron:

$ \begin{align*} E &= \text{Total energy absorbed} - \text{Threshold energy} \\ &= 3.27 \times 10^{-18} \, \text{J} - 2.18 \times 10^{-18} \, \text{J} \\ &= 1.09 \times 10^{-18} \, \text{J} \, \text{(ii)} \end{align*} $

The wavelength $\lambda$ of the emitted electron is given by:

$ \lambda = \frac{h}{\sqrt{2mE}} $

Substituting the given values into this equation:

$ \lambda = \frac{h}{\sqrt{2 \times 9 \times 10^{-31} \times 1.09 \times 10^{-18}}} = \frac{h \times 10^{24}}{\sqrt{1.962}} \, \text{m} $

Let's refine the explanation for clarity and professionalism:

The problem involves calculating the uncertainty in the position of a golf ball using the Heisenberg uncertainty principle. We are given:

The mass of the ball: $ m $ grams, which converts to $ m = 1 \times 10^{-3} $ kilograms.

The speed (velocity) of the ball: $ v = 50 \, \text{ms}^{-1} $.

The speed measurement has an uncertainty of $ 2\% $. Therefore, the uncertainty in speed, $ \Delta v $, is calculated as follows:

$ \Delta v = 2\% \text{ of } 50 \, \text{ms}^{-1} = \frac{2 \times 50}{100} = 1 \, \text{ms}^{-1} $

According to the Heisenberg uncertainty principle:

$ \Delta x \cdot \Delta p = \frac{h}{4 \pi} $

where $\Delta x$ is the uncertainty in position and $\Delta p$ is the uncertainty in momentum. Since momentum $ p $ is mass times velocity ($ p = m \times v $), we can write:

$ \Delta x \cdot m \Delta v = \frac{h}{4 \pi} $

Solving for $\Delta x$, we get:

$ \Delta x = \frac{h}{4 \pi m \Delta v} $

Substituting the known values, we find:

$ \Delta x = \frac{h}{4 \pi \times 1 \times 10^{-3} \times 1} = \frac{h}{4 \pi m} \times 10^3 \, \text{m} $

This result shows the uncertainty in the position is $\frac{h}{4 \pi m} \times 10^3$ meters.

To calculate the de-Broglie wavelength of a particle, we use the formula:

$ \lambda = \frac{h}{mv} $

where $ \lambda $ is the wavelength, $ h $ is Planck's constant, $ m $ is the mass of the particle, and $ v $ is the velocity of the particle.

Given:

Mass $ m = 1 \, \text{mg} = 1 \times 10^{-3} \, \text{g} = 1 \times 10^{-6} \, \text{kg} $

Velocity $ v = 10 \, \text{m/s} $

Planck's constant $ h = 6.63 \times 10^{-34} \, \text{Js} $

Substitute these values into the formula:

$ \lambda = \frac{6.63 \times 10^{-34}}{1 \times 10^{-6} \times 10} = \frac{6.63 \times 10^{-34}}{10^{-5}} = 6.63 \times 10^{-29} \, \text{m} $

Therefore, the de-Broglie wavelength of the particle is $ 6.63 \times 10^{-29} \, \text{m} $.

The correct set of four quantum numbers for the valence electron of strontium $(Z=38)$ is derived from its electronic configuration.

Electronic Configuration of Strontium $(Z=38)$:

$ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^6 \, 5s^2 $

For the Valence Shell Electron $(5s^2)$:

Principal Quantum Number ($n$): 5

Azimuthal Quantum Number ($l$): 0 (since it is an $s$-orbital)

Magnetic Quantum Number ($m$): 0 (since $m$ ranges from $-l$ to $+l$, and for $l=0$, $m=0$)

Spin Quantum Number: $\frac{1}{2}$

Thus, the correct set of quantum numbers for the valence electron is $(5, 0, 0, \frac{1}{2})$.

Energy of photon $\left(E_{\text {photon }}\right)=\frac{h c}{\lambda}$

$ =6.63 \times 10^{-34} \mathrm{Js} $

$c=$ Speed of light $=3 \times 10^8 \mathrm{~m} / \mathrm{s}$

$\lambda=$ Wavelength of light emitted $={ }^{\prime} \chi^{\prime} \mathop {\rm{A}}\limits^{\rm{o}} $

$ \begin{aligned} =x & \times 10^{-10} \mathrm{~m} \\ E_{\text {photon }} & =\frac{6.63 \times 10^{-34} \mathrm{Js} \times 3 \times 10^8 \mathrm{~m} / \mathrm{s}}{x \times 10^{-10} \mathrm{~m}} \\ & =\frac{19.89 \times 10^{-16}}{x} \mathrm{~J} \end{aligned} $

Now, total energy $=$ Number of photons × $E_{\text {photon }}$

$ \begin{aligned} 100 & =2.0 \times 10^{20} \mathrm{~s}^{-1} \times \frac{19.89 \times 10^{-16}}{x} \\ 100 & =\frac{397800}{x} \\ \Rightarrow \quad x & =3978 \end{aligned} $

For hydrogen atom, Energy of an electron, $E_n=-\frac{13.58}{n^5} \mathrm{eV}$

For 1st energy level,

$ E_1=\frac{-13.58}{1^2} \mathrm{eV}=-13.58 \mathrm{eV} $

For 2 nd energy level, $E_2$

$ \begin{aligned} & =\frac{-13.58}{2^2} \mathrm{eV}=-3.395 \mathrm{eV} \\ \therefore & E_2-E_1=10.19 \mathrm{eV} \end{aligned} $

For 3rd energy level,

$ \begin{aligned} & \quad E_3=\frac{-13.58}{3^2} \mathrm{eV}=-1.51 \mathrm{eV} \\ & E_3-E_2=1.89 \mathrm{eV} \\ & \therefore \text { Ratio of }\left(E_2-E_1\right) \text { to } \\ & \left(E_3-E_2\right)=\frac{10.19 \mathrm{eV}}{1.89 \mathrm{eV}}=\frac{27}{5} \end{aligned} $

Thus, the ratio of difference between 1st and 2nd Bohr orbit energy to that of between 2nd and 3rd orbit energy is $27 / 5$.

In reaction $\mathrm{I}, \mathrm{Na}_2\left[\mathrm{Be}(\mathrm{OH})_4\right]$ is formed.

∴ Covalency of Be in $\mathrm{Na}_2\left[\mathrm{Be}(\mathrm{OH})_4\right]$ is 4 . In reaction II, $\mathrm{Na}\left[\mathrm{Al}(\mathrm{OH})_4\right]$ is formed.

∴ Covalency of AI in the complex is 4 .

The energy of $\mathrm{Li}^{2+}$ ion can be given as

$ E=-2.18 \times 10^{-18}\left(\frac{Z^2}{n^2}\right) \mathrm{J} / \text { atom } $

For $\mathrm{Li}^{2+}, Z=3$

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, n=1 $

So, $\quad E_1=-2.18 \times 10^{-18}\left[\frac{(3)^2}{(1)^2}\right] \mathrm{J} /$ atom

$ \begin{array}{r} E_1=-19.62 \times 10^{-18} \mathrm{~J} / \text { atom } \\ -1.962 \times 10^{-17} \mathrm{~J} \end{array} $

$Z=24$

$ C_{24}=1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^1 3 d^5 $

Last electrons enters in $4 s$ orbital.

(i) Hence, for $(n+l)=3$

Possible combination from electronic configuration are $2 p, 3 s$. Total electron $=6+2=8$

(ii) For $(n+l)=4$

Possible combinations are $3 p^6, 4 s^1$.

Total electrons $=6+1=7$

(iii) For $(n+l)=5$

Possible combination are $=3 p^5$

Total electrons $=5$

According to Bohr's theory,

Angular momentum, $m v r=\frac{n h}{2 \pi}$

For hydrogen atom, the value of $n$ in ground state $=1$

$ \begin{aligned} \therefore \text { Angular momentum } & =\frac{1 \times 6.625 \times 10^{-34}}{2 \times 3.14} \\ & =1.055 \times 10^{-34} \mathrm{Js}=1055 \times 10^{-37} \mathrm{Js} \end{aligned} $

The energy of the electron in the $n$th orbit of hydrogen $n$-atom is given by

$ \begin{aligned} \left(E_n\right) & =-R_{\mathrm{H}}\left(\frac{1}{n^2}\right) ; \quad E_1=-R_{\mathrm{H}}\left(\frac{1}{1}\right) \\ E_2 & =-R_{\mathrm{H}}\left(\frac{1}{4}\right) ; \quad E_3=-R_{\mathrm{H}}\left(\frac{1}{9}\right) \\ \therefore E_1: E_2: E_3 & =\frac{1}{1}: \frac{1}{4}: \frac{1}{9}=36: 9: 4 \end{aligned} $

Bohr's radius of $n$th orbit,

$ r_n=0.529\left[\frac{n^2}{Z}\right] \mathop {\rm{A}}\limits^{\rm{o}} $

where, $n=$ Energy level

$Z=$ Atomic number

For 1st Bohr orbit of H-atom, $n=1$ and $Z=1$

Radius of H -atom, $r_{\mathrm{H}}=0.529 \mathop {\rm{A}}\limits^{\rm{o}} $

For $\mathrm{He}^{+}$ion, $n=2, Z=2$

$ r_{\mathrm{He}^{+}}=0.529 \times\left[\frac{2^2}{2}\right]=2 r_{\mathrm{H}} $

For $\mathrm{Li}^{2+}$ ion, $n=2, Z=3$

$ r_{\mathrm{L}^{2+}}=0.529 \times\left[\frac{2^2}{3}\right]=\frac{4}{3} r_{\mathrm{H}} $

For $\mathrm{Be}^{3+}$ ion, $n=2, \mathrm{Z}=4$

$ r_{\mathrm{Be}^{3+}}=0.529 \times\left[\frac{2^2}{4}\right]=r_{\mathrm{H}} $

For $\mathrm{Be}^{3+}(n=2)$ has same radius as that of 1st Bohr's orbit of H -atom. So, $n=2$ and $X=\mathrm{Be}^{3+}$.

$n=3, l=3, m=1, s=\frac{1}{2}$ is impossible set of quantum numbers. When the principal quantum number ( $n$ ) is 3 , the possible azimuthal quantum number will be only 0,1 and 2 which represent $s, p$ and $d$ orbitals respectively.

Hybridisation $=\frac{1}{2}[V+X-C+A]$

$V=$ No. of valence electrons

$X=$ No. of monovalent atoms around central atom

$C=$ Positive charge on cation

$A=$ Negative charge on anion

$ \begin{array}{cccccc} \hline \text { Molecules } & \boldsymbol{V} & \boldsymbol{X} & \boldsymbol{C} & \boldsymbol{A} & \text { Hybridisation } \\ \hline \mathrm{BF}_3 & 3 & 3 & 0 & 0 & \frac{1}{2}[3+3]=3 \\ & & & & & =s p^2 \\ \hline \mathrm{BeF}_2 & 2 & 2 & 0 & 0 & \frac{1}{2}[2+2]=2 \\ & & & & & =s p \\ \hline \mathrm{BrF}_3 & 7 & 3 & 0 & 0 & \frac{1}{2}[7+3]=5 \\ & & & & & =s p^3 d \\ \hline \end{array} $

Alkali metals have low ionisation energy as compared to alkaline earth metals due to their large size and their valence shell electronic configuration. Alkali metals easily loses it's one electron to attain noble gas configuration, which is stable.

If element behave as reducing agent, it means it undergoes oxidation. If element behave as oxidising agent, it means it undergoes reduction.

$\mathrm{Cr}^{2+}$ has $3 d^4$ configuration. It will lose electron to form $3 d^3$ configuration, which is stable (as it has half-filled $t_{2 g}$ level). Hence, it is a reducing agent. $\mathrm{Mn}^{3+}$ also has $3 d^4$ configuration. It will gain electron to form $\mathrm{Mn}^{2+}$, which is stable $3 d^5$ configuration. Hence, it is an oxidising agent.

For the Balmer series of the hydrogen atom, we use the formula:

$ \frac{1}{\lambda} = R_{\mathrm{H}}\left[\frac{1}{2^2} - \frac{1}{n^2}\right] $

For the second line of the Balmer series, $ n = 4 $. Therefore, the equation simplifies to:

$ \begin{aligned} \frac{1}{\lambda} &= R_{\mathrm{H}}\left[\frac{1}{4} - \frac{1}{16}\right] \\ &= \frac{3 R_{\mathrm{H}}}{16} \\ \lambda &= \frac{16}{3 R_{\mathrm{H}}} \end{aligned} $

Next, for the Lyman series of the $\mathrm{He}^{+}$ ion, the formula is:

$ \frac{1}{\lambda_{\mathrm{He}^{+}}} = 4 R_{\mathrm{H}}\left[\frac{1}{1^2} - \frac{1}{n^2}\right] $

For the first line of the Lyman series, $ n = 2 $. Substituting this value gives:

$ \begin{aligned} \frac{1}{\lambda_{\mathrm{He}^{+}}} &= 4 R_{\mathrm{H}}\left[1 - \frac{1}{4}\right] \\ &= 4 R_{\mathrm{H}} \cdot \frac{3}{4} \\ &= 3 R_{\mathrm{H}} \end{aligned} $

Thus, calculating the wavelength:

$ \lambda_{\mathrm{He}^{+}} = \frac{1}{3 R_{\mathrm{H}}} $

We now find the relationship between $\lambda_{\mathrm{He}^{+}}$ and $\lambda$:

$ \begin{aligned} \frac{\lambda_{\mathrm{He}^{+}}}{\lambda} &= \frac{1}{3 R_{\mathrm{H}}} \times \frac{3 R_{\mathrm{H}}}{16} \\ \lambda_{\mathrm{He}^{+}} &= \frac{\lambda}{16} \end{aligned} $

Therefore, the wavelength of the first line of the Lyman series for the $\mathrm{He}^{+}$ ion is $\frac{\lambda}{16}$.

The correct statements are II and IV only. While statements I and III are incorrect.

The direction of spin is described by spin quantum number. This quantum number gives information about the direction of spinning of electron present in any orbit.

Hence, statement I is incorrect.

Energy of photon is given by $E=\frac{h c}{\lambda}$

$\therefore$ Energy is inversely proportional to wavelength ( $\lambda$ ).

Atso, $E=h c \bar{v}$

$\therefore$ Energy is directly proportional to wave number $(v)$. So, statement III is also incorrect.

Let’s work through the problem step by step.

For hydrogen-like systems, the radius of the nth orbit is given by:

$r_n = \frac{n^2}{Z} a_0$

where:

$n$ is the principal quantum number,

$Z$ is the atomic number (nuclear charge), and

$a_0$ is the Bohr radius.

For the hydrogen atom (with $Z = 1$) in the third orbit:

$r_3 = 3^2 \cdot a_0 = 9a_0$

We are given that:

$9a_0 = R$

Hence, the Bohr radius in this context is:

$a_0 = \frac{R}{9}$

For the $\mathrm{He}^+$ ion, the nucleus has $Z = 2$. For the second orbit ($n = 2$), the radius is:

$r_2 = \frac{2^2}{2} a_0 = \frac{4}{2} a_0 = 2a_0$

Substitute the value of $a_0$ into the expression for $r_2$:

$r_2 = 2 \left(\frac{R}{9}\right) = \frac{2}{9} R$

Thus, the radius of the second orbit of the $\mathrm{He}^+$ ion is $\frac{2}{9} R$.

The correct answer is Option D.

Given:

Threshold frequency of the metal, $ v_0 = 10^{15} \, \text{s}^{-1} $

First frequency, $ v_1 = 1.5 \times 10^{15} \, \text{s}^{-1} $

Second frequency, $ v_2 = 2.0 \times 10^{15} \, \text{s}^{-1} $

We need to find the ratio of the maximum kinetic energies of the photoelectrons, $\frac{(\text{KE}_{\max})_1}{(\text{KE}_{\max})_2}$.

Using the photoelectric effect equation:

$ hv = hv_0 + \text{KE} $

Therefore, the kinetic energy (KE) is given by:

$ \text{KE} = h(v - v_0) $

The ratio of the maximum kinetic energies is calculated as follows:

$ \frac{(\text{KE}_{\max})_1}{(\text{KE}_{\max})_2} = \frac{h(v_1 - v_0)}{h(v_2 - v_0)} $

Substitute the given values:

$ \frac{(\text{KE}_{\max})_1}{(\text{KE}_{\max})_2} = \frac{h(1.5 \times 10^{15} - 1 \times 10^{15})}{h(2 \times 10^{15} - 1 \times 10^{15})} $

Simplifying the expression:

$ \frac{\text{KE}_1}{\text{KE}_2} = \frac{0.5 \times 10^{15}}{1 \times 10^{15}} $

Thus:

$ \frac{\text{KE}_1}{\text{KE}_2} = \frac{1}{2} \quad \text{or} \quad \text{KE}_1 : \text{KE}_2 = 1:2 $

(A) There is no time lag between the striking of light and ejection of electrons from the metal surface. The statement is correct as the electron gets immediately ejected as soon as the light of appropriate frequency strikes the metal surface.

(B) The number of electrons ejected is independent of the intensity of light is incorrect. The number of electrons ejected depends on the intensity of incident radiation.

(C) The elements $\mathrm{K}, \mathrm{Rb}$ and Cs can show photoelectric effect when exposed to the beam of light is correct. Photoelectric effect can be normally observed in alkali metals as these have low incident energy value/work function.

$\therefore(A)$ and $(C)$ both are correct.

In $n=4$ energy level, $l=0,1,2,3$ i.e. $s, p, d, f$ subshells.

Magnetic quantum number $m_l$ have values from

$ -l \text { to }+l . $

For $n=4$, possible values of $l$ and $m_l$ are

$ \begin{aligned} & l=0, m_l=0 \\ & l=1, m_l=-1,0,+1 \\ & l=2, m_l=-2,-1,0,+1,+2 \\ & l=3, \\ & m_l=-3,-2,-1,0,+1,+2,+3 \end{aligned} $

Total number of orbitals $=16$

Total number of electrons will be 32.

∴ Maximum number of electrons with spin value $\left(+\frac{1}{2}\right)=16$

Speed of light in vacuum

$ =3 \times 10^8 \mathrm{~m} / \mathrm{s} . $

Velocity of electron in first Bohr's orbit of H -atom is

$ \begin{aligned} V_n & =2.18 \times 10^6 \times \frac{Z}{n} \mathrm{~m} / \mathrm{s} \\ & =2.18 \times 10^6 \quad(Z=1, n=1) \end{aligned} $

Ratio of speed of light in vacuum to speed of electron in first Bohr orbit of H -atom is $3 \times 10^8: 2.18 \times 10^6$

$137: 1$.

Radius of Bohr's orbit in hydrogen atom can be calculated as

$ \begin{aligned} & r=\frac{n^2 h^2}{4 \pi^2 m e^2} \times \frac{1}{Z} \\ or\,\,\,\,& r=0.529 \times \frac{n^2}{Z} \mathop {\rm{A}}\limits^{\rm{o}} \end{aligned} $

where, $n=$ number of the orbit

$ Z=\text { atomic number } $

For hydrogen's third Bohr's orbit,

$ \begin{aligned} r_3 & =0.529 \times \frac{(3)^2}{1} \mathop {\rm{A}}\limits^{\rm{o}} \\ & =9 \times 0.529 \mathop {\rm{A}}\limits^{\rm{o}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} $

Also $\quad r_2=0.529 \times(2)^2$

(Given, radius of second Bohr's orbit radius.)

$ \begin{aligned} r_2 & =4 \times 0.529 \\ r_2 / 4 & =0.529 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

From Eqs. (i) and (ii),

$ \begin{aligned} r_3 & =9 \times \frac{r_2}{4} \\ \Rightarrow \quad r_3 & =\frac{9}{4} r_2 \end{aligned} $

Hence, radius of third Bohr's orbit

$ =\frac{9}{4} r_2 $

Energy of $n$th Bohr's orbit is given as

$ E_n=\frac{k Z^2}{n^2} $

where, $n=$ number of orbit,

$Z=$ atomic number [For

hydrogen, $Z=1$ ]

$ \therefore \quad E_2=k \frac{(1)^2}{(2)^2} $

[Given, second Bohr's orbit energy.]

$ \begin{aligned} E_2 & =k / 4 \\ k & =E_2 \times 4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

Also, $\quad E_3=k \frac{(1)^2}{(3)^2}=\frac{k}{9}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

From Eqs. (iii) and (iv),

$ E_3=\frac{E_2 \times 4}{9} \Rightarrow E_3=\frac{4}{9} E_2 $

Hence, energy of third Bohr's orbit is $\frac{4}{9} E_2$.

For the occurrence of photoelectric effect, a minimum amount of energy, i.e. equal to work function $(\phi)$ is to be given to an electron.

The metals having lower values of work function than the incident energy values will show photoelectric effect. Incident energy can be calculated as

$ E=\frac{h c}{\lambda} $

$ \text { } \begin{aligned} where,\,\, E & =\text { energy, } \\ h & =\text { Planck's constant } \\ \qquad & \left(6.63 \times 10^{-34} \mathrm{Js}\right) \\ c & =\text { speed of light }\left(3 \times 10^8 \mathrm{~m} / \mathrm{s}\right) \\ \lambda & =\text { wavelength }(300 \mathrm{~nm}) \\ E & =\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9} \mathrm{~m} \times 1.6 \times 10^{-19}} \mathrm{eV} \\ & =4.14 \mathrm{eV} \end{aligned} $

$ \begin{array}{|l|l|l|l|l|l|} \hline \text { Metal } & \mathrm{Li} & \mathrm{Mg} & \mathrm{Ag} & \mathrm{Cu} & \mathrm{~K} \\ \hline(\phi) \text { eV } & 2.4 & 3.7 & 4.3 & 4.8 & 2.2 \\ \hline \end{array} $

Here, we can see that $\mathrm{Li}, \mathrm{Mg}$ and K have lower value of $\phi$ than incident energy. Therefore, they will show photoelectric effect.

Given,

$ \Delta v=\frac{1}{2 m} \sqrt{\frac{h}{\pi}} $

Using equation,

$ \Delta x \cdot \Delta v=\frac{h}{4 m \pi} $

Dividing both sides by $m \Delta v$;

$ \frac{\Delta x \cdot \Delta v}{m \Delta v}=\frac{h}{4 m^2 \pi \Delta v} $

As,  $ \Delta \rho=m \Delta v $

$ So,\frac{\Delta x}{\Delta \rho}=\frac{h}{4 m^2 \pi \Delta v^2} $

Put the value of $\Delta v$ from Eq. (i)

$ \begin{aligned} & \frac{\Delta x}{\Delta \rho}=\frac{h \times m^2 \times 4 \times \pi}{4 m^2 \pi h} \\ \therefore & \frac{\Delta x}{\Delta \rho} =\frac{1}{1}\end{aligned} $

The ratio of uncertainty in position and momentum is $1: 1$.

Given, $n_1=2 ; n_2=6$

Total number of spectral lines when electron returns from 6th shell to 2nd

$ \begin{aligned} \text { atom } & =\frac{\left(n_2-n_1\right)\left(n_2-n_1+1\right)}{2} \\ & =\frac{(6-2)(6-2+1)}{2}=\frac{4 \times 5}{2}=10 \end{aligned} $

For an electron in $d$-orbital, $l=2$

orbital angular momentum

$ =\sqrt{l(l+1)} \hbar=\sqrt{l(l+1)} \frac{\hbar}{2 \pi} $

Now, $l=2 \Rightarrow \sqrt{2(2+1)} \hbar$

$ =\sqrt{2(2+1)} \hbar\left\{\hbar=\frac{h}{2 \pi}\right\}=\sqrt{6} \hbar $

=Energy required to transfer an electron in hydrogen atom is given by,

$ \Delta E=\frac{-13.6}{n_2^2}-\frac{-13.6}{n_1^2} $

Here, $n_1=2$ and $n_2=3$

$ E_{\mathrm{H}}=\frac{-13.6}{3^2}-\frac{-13.6}{2^2}=1.89 \mathrm{eV} \simeq 1.9 \mathrm{eV} $

We know that, $\lambda=\frac{h}{m v}$

(where, $v$ is velocity.)

So, $\quad v=h / m \lambda$ (where, $v$ is frequency.) and $\lambda=C / v$

In relation, $m v=h / \lambda$, we can substitute the value of $\lambda$.

$ m v=\frac{h}{C / v} \Rightarrow v=\frac{h v}{m C} $

The wavelength associated with an electron moving in the first orbit of a hydrogen atom with a velocity of $2.2 \times 10^6 \, \mathrm{ms}^{-1}$ (in nm) can be determined as follows:

Given values:

$m_e = 9.0 \times 10^{-31} \, \mathrm{kg}, \, h = 6.626 \times 10^{-34} \, \mathrm{Js}$

According to de Broglie's equation:

$\lambda = \frac{h}{mv}$

Where:

$\lambda$ is the wavelength,

$h$ is Planck's constant ($6.626 \times 10^{-34} \, \mathrm{Js}$),

$m$ is the mass of the electron ($9.0 \times 10^{-31} \, \mathrm{kg}$),

and $v$ is the velocity of the electron ($2.2 \times 10^6 \, \mathrm{m/s}$).

Substituting the given values:

$ \begin{aligned} \lambda &= \frac{6.626 \times 10^{-34}}{9.0 \times 10^{-31} \times 2.2 \times 10^6} \\ &= 0.33 \times 10^{-9} \, \mathrm{m} \end{aligned} $

This simplifies to 0.33 nm (since $1 \, \mathrm{nm} = 10^{-9} \, \mathrm{m}$).

The energy required (in eV) to excite an electron in a hydrogen atom from the ground state to the third state is calculated as follows:

First, we use the formula for the energy of an electron in a specific state:

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

For the ground state (n = 1) of a hydrogen atom (Z = 1):

$ E_1 = \frac{-13.6 \times 1^2}{1^2} = -13.6 \text{ eV} $

For the third state (n = 3) of a hydrogen atom (Z = 1):

$ E_3 = \frac{-13.6 \times 1^2}{3^2} = -1.511 \text{ eV} $

The energy required to excite the electron from the ground state to the third state is given by:

$ \Delta E = E_3 - E_1 = -1.511 - (-13.6) = 12.089 \text{ eV} \approx 12.1 \text{ eV} $

An orbital can have maximum two electrons. Hence, an orbital with $n=4$ and $l=3$ can have maximum two electrons.

Azimuthal or angular quantum number ($l$) describes about the shape of the orbitals. Orbital having $l=0$ have spherical shape $l=1$, Dumbell $l=2$, Double dumbell

Given, $\Delta x=$ uncertainty in position

$\Delta v=$ uncertainty in velocity we know, $\Delta p=m \Delta v$, where $\Delta p$ is the uncertainty in momentum and $m$ is the mass of particle. Using Heisenberg's uncertainty principle,

$\qquad\Delta x \cdot m \Delta v=\frac{h}{4 \pi}$

As $\quad \Delta x=\Delta v$

So, $\Delta v \cdot m \Delta v=\frac{h}{4 \pi}$

On multiplying both sides with $m$, we get

$m \Delta v \cdot m \Delta v=\frac{m h}{4 \pi}$

Thus, $\Delta p^2=\frac{m h}{4 \pi} \Rightarrow \Delta p=\sqrt{\frac{m h}{4 \pi}}$

Hence, $\Delta p=\frac{1}{2} \sqrt{\frac{m h}{\pi}}$

Number of radial nodes, $R_n=n-l-1$

Number of angular nodes, $A_n=l$

For $4 f$-orbital,

Principal quantum number, $n=4$

Azimuthal quantum number, $l=3$

Thus,

$\begin{aligned} & R_n=4-3-1=0 \\ & A_n=3 \end{aligned}$

Hence, radial nodes and angular nodes for $4 f$-orbital are 0 and 3 respectively.

$n=3$, means $3 \mathrm{rd}$ shell

Azimuthal quantum number, $l=n-1$

$\Rightarrow \quad l=0,1,2$

(0 means $s$ subshell, 1 means $p$ subshell and 2 means $d$ subshell)

$\text { So, } \quad l=2 \text { means } d \text { subshell }$

$\therefore 3 d$ subshell can accommodate maximum $2(2 l+1)$ 10 electrons.

$\begin{aligned} & \text { } \psi(\text { work function })=\frac{h c}{\lambda_0} \\ & \lambda_0=\text { threshold wavelength } \\ & 3.75 \times 1.6 \times 10^{-19}=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{\lambda_0} \\ & \lambda_0=\frac{6.62 \times 3 \times 10^{-34} \times 10^8}{3.75 \times 1.6 \times 10^{-19}} \\ & =3.30 \times 10^{-7}=330 \times 10^{-9} \\ & =330 \mathrm{~nm} \text {. } \\ & \end{aligned}$

Difference in energy is given as

$\begin{aligned} \Delta E & =\frac{-Z^2 R}{n^2} \\ \Delta E & \propto \frac{1}{n^2} \end{aligned}$

Therefore, as principal quantum number $(n)$ increases, the energy of transition decreases (Negative sign indicates the spin of electron).

Number of protons, neutrons and electrons in $_6^{13}$C are

Atomic number = 6

Atomic number = number of protons = number of electrons = 6

$\therefore$ Protons (p) = 6

Electrons (e) = 6

Atomic mass = number of protons + number of neutrons

13 = 6 + x $\Rightarrow$ x = 7 (Atomic mass = 13)

Mass of proton $(p)=1.6726 \times 10^{-27} \mathrm{~kg}$

Mass of neutron $(n)=1.6749 \times 10^{-27} \mathrm{~kg}$

Mass of electron $(e)=9.1 \times 10^{-31} \mathrm{~kg}$

Mass of proton is 1836.16 times heavier than electron.

Mass of neutron is 1838.68 times heavier than electron.

Ratio of mass of electron, proton and neutron will be

Mass of $e:$ Mass of $p:$ Mass of $n$

$1: 1836.16: 1838.68$

Number of electrons present in the following species

Atomic number $=$ number of electrons in neutral species.

(i) $\mathrm{Be}^{2+} \Rightarrow$ Atomic number $=4$

It donate 2 electrons and acquire +2 charge.

$\therefore$ Number of electrons $=2$

(ii) $\mathrm{H}^{+} \Rightarrow$ Atomic number $=1$

It donate 1 electron and acquire +1 charge.

Number of electrons $=0$

(iii) $\mathrm{Na}^{+} \Rightarrow$ Atomic number $=11$

+1 charge represent donation of 1 electron.

Number of electrons $=10$

(iv) $\mathrm{Mg}^{+} \Rightarrow$ Atomic number $=12$

$\mathrm{Mg}^{+}$ represent donation of 1 electron.

Number of electrons $=11$

$\begin{aligned} \therefore \quad \mathrm{Be}^{2+} & =2, \mathrm{H}^{+}=0 \\ \mathrm{Na}^{+} & =10, \mathrm{Mg}^{+}=11 \end{aligned}$

Metals emit electrons when light shines upon them. The phenomenon is called as photoelectric effect and the electrons emitted in such manner are called photoelectrons.

Relation between kinetic energy and frequency is kinetic energy $=\phi-h \mathrm{v}$

Here $\phi=$ metal work function

Kinetic energy of photoelectrons depends on frequency of incident light and metal work function. $\mathrm{Cu}$ has maximum metal work function's value.

Hence, $\mathrm{Cu}$ have minimum kinetic energy of ejected photoelectrons than alkali metals.

We know that, energy $=h c / \lambda$

Here, $h=$ Planck's constant (Js )

$c=$ speed of light $(\mathrm{m} / \mathrm{s})$

$\lambda=$ wavelength $(\mathrm{m})$

Now for energy $E_1=25 \mathrm{eV}$

$E_1=h c / \lambda_1 \Rightarrow 25=h c / \lambda_1$ ..... (i)

Similarly for energy,

$E_2=h c / \lambda_2$

$100=h c / \lambda_2$ ..... (ii)

By divide $E_1$ and $E_2$, we get

$\begin{aligned} & 25 / 100=\lambda_2 / \lambda_1 \Rightarrow 1 / 4=\lambda_2 / \lambda_1 \\ & \lambda_1: \lambda_2=4: 1 \text { or } \lambda_1=4 \lambda_2 \end{aligned}$

Let us consider, particle $A$ having velocity $=v_A$ and particle $B$ having velocity $=v_B$

According to de-Broglie wavelength, $\lambda=\frac{h}{m v}$

here, $h=$ Planck constant

$m=$ mass of electron

$v=$ velocity of particle

Now, $\lambda_A=\frac{h}{m_A v_A}$ and $\lambda_B=\frac{h}{m_B v_B}$

Wavelength of $A$ is found to be double than of $B$.

$\text { i.e. } \lambda_A=2 \times \lambda_B$ ..... (i)

Put value of $\lambda_A$ and $\lambda_B$ in Eq. (i)

$\begin{array}{rlrl} 2 \times \frac{h}{m_B v_B} & =\frac{h}{m_A v_A} & & \left(\text { For } v_A=v_B\right) \\ \frac{2}{m_B} & =\frac{1}{m_A} & {\left[m_A=\frac{m_B}{2}\right]}\end{array}$

Here, mass of $A$ is half that of $B$.