Atomic Structure

$ \begin{aligned} &\\ &\begin{aligned} \text { } \lambda & =221 \mathrm{~nm}=221 \times 10^{-9} \mathrm{~m} \\ \text { Using } E & =\frac{h c}{\lambda}=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{221 \times 10^{-9}} \\ E & =9 \times 10^{-19} \mathrm{~J} \\ \mathrm{KE}_{\max } & =E-\phi \\ \Rightarrow \mathrm{KE}_{\max } & =\left(9.0 \times 10^{-19}-7.68 \times 10^{-19}\right) \\ \mathrm{KE}_{\max } & =1.32 \times 10^{-19} \mathrm{~J} \end{aligned} \end{aligned} $

$Z=25=1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^5 4 s^2$

For $2 p^6: n=2, l=1$, so $(n+l)=3$

It contains 6 electrons

For $3 s^2: n=3, l=0,(n+l)=3+0=3$

If contains 2 electrons

$ x=6+2=8 $

For $3 p^6: n=3, l=1,(n+l)=4$

It contains 6 electrons

For $4 s^2, n=4, l=0,(n+l)=4$

It contains 2 electrons

$ \begin{aligned} & Y=2+6=8 \\ & x+y=8+8=16 \end{aligned} $

Using the Rydberg formula

$ \frac{1}{\lambda}=R Z^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right] $

For a given transition, $\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$ is constant.

So $\frac{1}{\lambda} \propto Z^2, \lambda \propto \frac{1}{Z^2}$

Therefore, $\lambda_{\mathrm{H}} Z_{\mathrm{H}}^2=\lambda_{\mathrm{He}^{+}} Z_{\mathrm{He}^{+}}^2$

Substituting the values,

$ \begin{aligned} & \lambda_{\mathrm{H}} \times(1)^2=100 \mathrm{~nm} \times(2)^2 \\ & \lambda_{\mathrm{H}}=400 \mathrm{~nm} \text { or } 4000 \mathop {\rm{A}}\limits^{\rm{o}} \end{aligned} $

Energy of electron in $n$th orbit

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

Energy of 2nd orbit of hydrogen,

$ E_2=\frac{-1^2}{2^2} \times 13.6 \mathrm{eV}=-3.4 \mathrm{eV} $

For $\mathrm{He}^{+}, n=4$

$ E_4=-\frac{2^2}{4^2} \times 13.6 \mathrm{~V}=-3.4 \mathrm{eV} $

First of all calculate the frequency of $c$ and $d$

$ \begin{aligned} v_c & =\frac{c}{\lambda_c}=\frac{3 \times 10^8}{400 \times 10^{-9}} \\ & =7.5 \times 10^{14} \mathrm{~Hz} \\ v_d & =\frac{c}{\lambda_d}=\frac{3 \times 10^8}{750 \times 10^{-8}} \\ & =4 \times 10^{14} \mathrm{~Hz} \end{aligned} $

Now we know

$ E=h v \text { or } E \propto v $

So, higher the frequency, greater will be the energy thus, $E$ is in order

$ \begin{aligned} 3 \times 10^{15} \mathrm{~Hz}>7.5 \times 10^{14} \mathrm{~Hz} & >4 \times 10^{14} \mathrm{~Hz} \\ & >2 \times 10^{14} \mathrm{~Hz} \end{aligned} $

or the correct increasing order is $b

$\operatorname{Cr}(24)=[\operatorname{Ar}] 3 d^5 4 s^1$

For $1 s^2, 2 s^2, 3 s^2, 4 s^1$ each $s$-orbital has $m_l=0$

contibuting 2 electron (except $4 s^1$ )

For $2 p^6, 3 p^6$ : each $p$ shell has one orbital $m_l=0$

Contributing 2 electrons each

For $3 d^5: d$ subshell has one orbital with $m_l=0$

Contributing 1 electron.

Total $2+2+2+1+2+2+1=12$

$\mathrm{Cu}=[\mathrm{Ar}] 3 d^{10} 4 s^1$

$1 s^2 2 s^2 3 s^2 4 s^1=$ each will contribute 2 electrons

$2 p^6, 3 p^6$ : each will contribute 2 electrons

$3 d^{10}$ : contribute 2 electron

$ 2+2+2+1+2+2+2=13 $

Using Rydberg formula

$ \frac{1}{\lambda}=R z^2\left[\frac{1}{n_f^2}-\frac{1}{n_i^2}\right] $

for $\mathrm{He}^{+}, z=2$, for Balmer series, $n_f=2$

The initial energy level, $n_i=n$ where $n>2$

$ \begin{aligned} \frac{1}{\lambda} & =R(2)^2\left[\frac{1}{2^2}-\frac{1}{n^2}\right] \\ \Rightarrow \quad \frac{1}{\lambda} & =4 R\left(\frac{1}{4}-\frac{1}{n^2}\right) \end{aligned} $

On simplifying above equation

$ \begin{aligned} \lambda & =\frac{n^2}{R\left(n^2-4\right)} \\ \Rightarrow \quad \lambda & =\frac{n^2}{R(n-2)(n+2)} \end{aligned} $

Using the formula, $E=\frac{h c}{\lambda}$

The approx value of $h c=120 \mathrm{eV} . \mathrm{nm}$

$ E=\frac{1240 \mathrm{eV} \cdot \mathrm{~nm}}{300 \mathrm{~nm}}=4.13 \mathrm{eV} $

For $\mathrm{Mg}: 4.13 \mathrm{eV}>3.7 \mathrm{ev}$, so electron will be ejected

For $\mathrm{Cu}: 4.13 \mathrm{eV}<4.8 \mathrm{eV}$, no electron ejected

For $\mathrm{Ag}: 4.13<4.3 \mathrm{eV}$, no electron ejected

For $\mathrm{Na}: 4.13>2.3 \mathrm{eV}$, so electron will be ejected

Thus, 2 metals will eject electron.

Uncertainty principle for particle $A$;

$ \Delta x_A\left(m_A \cdot \Delta v_A\right)=\frac{h}{4 \pi} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

For particle $B$,

$ \Delta x_B\left(m_B \cdot \Delta v_B\right)=\frac{h}{4 \pi} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Divide Eq. (i) by Eq. (ii)

$ \frac{\Delta x_A}{\Delta x_B}=\frac{m_B \cdot \Delta v_B}{m_A \cdot \Delta v_A} $

Now substitute the given value

$ \begin{aligned} m_B & =4 m_A, \Delta v_A=0.03 \\ \Delta v_B & =0.01 \\ \frac{\Delta x_A}{\Delta x_B} & =\frac{0.04}{0.03}=\frac{4}{3} \end{aligned} $

For $s$-orbital, $l=0$

For $d$-orbital, $l=2$

For $f$-orbital, $l=3$

For, $f$-orbital $m_1$ values can be $-3,-2,-1,0+1,+2,+3$

So, $4 f$ can have $m_l=-2$

Each orbital with $m_l=-2$ can hold 2 electrons.

Orbitals that can have $m_l=-2$ are $3 d, 4 f$ and $6 d$

Total electron $=2+2+2=6$

Using the formula for $n$th orbit of Bohr's orbit,

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

For $R_1$ of $\mathrm{He}^{+}$,

Given, $n=4, Z=2$

$ \begin{aligned} R_1 & =\frac{a_0 n^2}{Z} \\ \Rightarrow \quad R_1 & =52.9 \times \frac{16}{2}=423.2 \mathrm{pm} \end{aligned} $

For $R_2\left(\mathrm{Li}^{2+}\right)$

$ \begin{aligned} n & =3, z=3 \\ R_2 & =52.9 \times \frac{3^2}{3}=158.7 \mathrm{pm} \\ R_1-R_2 & =423.2-158.7=264.50 \mathrm{pm} \end{aligned} $

$ \begin{aligned} &\text { de-Broglie wavelength is, } \lambda=\frac{h}{p}\\ &\begin{aligned} \Rightarrow \quad \lambda & =\frac{h}{\sqrt{2 m K E}} \text { or } \quad \mathrm{KE}=\frac{h^2}{2 m \lambda^2} \\ \frac{(\mathrm{KE})_X}{(\mathrm{KE})_Y} & =\frac{\frac{h^2}{2 m \lambda_x^2}}{\frac{h^2}{2 m \lambda_Y^2}}=\frac{9}{9}=1: 1 \end{aligned} \end{aligned} $

The radii of $n$th orbit of hydrogen atom,

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

For hydrogen atom the radii of 3 rd and 2nd obit respectively can be calculated as

$ \begin{aligned} & r_3=\frac{a_0 3^2}{1}=9 a_0 \quad(z=1) \\ & r_2=\frac{a_0 2^2}{1}=4 a_0 \\ & x=r_3-r_2=9 a_0-4 a_0=5 a_0 \end{aligned} $

For $\mathrm{Li}^{2+}$ ion, the radii of 3rd and 4th orbit respectively can be calculated as

$ \begin{aligned} r_4 & =a_0 \frac{4^2}{3}=\frac{16 a_0}{3} \quad(z=3) \\ r_3 & =a_0 \frac{3^2}{3}=3 a_0 \\ y & =r_4-r_3=\frac{7 a_0}{3} \\ y: x & =\frac{7 a_0}{3} \times \frac{1}{5 a_0}=7: 15 \end{aligned} $

Radius of $n$th orbit of H -atom.

$ \begin{aligned} & r_n=r_0 \times n^2 \\ & r_3=3^2 \times 52.9 \mathrm{pm} \end{aligned} $

Using Bohr's postulate of angular momentum

$ m v r=\frac{n h}{2 \pi} \Rightarrow m v=\frac{3 h}{2 \pi r} $

According to de-Broglie

$ \begin{aligned} & \lambda=\frac{h}{m v}=\frac{h}{3 h / 2 \pi r}=\frac{2 \pi r}{3} \\ \Rightarrow & \lambda=\frac{2 \pi \times 3^2 \times 52.9}{3} \\ \Rightarrow & \lambda=6 \pi \times 52.9 \mathrm{pm} \end{aligned} $

Using the Rydberg formula for wavenumber

$ \bar{v}=R z^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right] $

for wavenumber of 1st line of H atom

$ \begin{aligned} n_1=2, n_2 & =3, z=1 \\ \bar{v}_{\mathrm{H}} & =R\left[\frac{1}{4}-\frac{1}{9}\right] \\ \bar{v}_{\mathrm{H}} & =R\left[\frac{5}{36}\right] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

The wavenumber of 2 nd line of $\mathrm{He}^{+}$

$ \begin{aligned} & n_1=2, n_2=4, z=2 \\ & \bar{v}_{\mathrm{He}}^{+}=4 R\left[\frac{1}{4}-\frac{1}{16}\right]=R\left(\frac{3}{4}\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Divide Eq. (ii) by Eq. (i)

$ \bar{v} \mathrm{He}^{+}=\frac{27}{5} \bar{v}_{\mathrm{H}} \text { or } \frac{27}{5} \bar{v}_1 $

For set $n=3, l=3, m-3, s=+\frac{1}{2}$ is not possible.

The value of $l$ varies from 0 to $(n-1)$

So, $n=3, l$ can have value $0,1,2$ only

Heisenberg's uncertainty principle,

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

Substituting the values

$ \begin{aligned} & \Delta p \geq \frac{6.626 \times 10^{-34} \mathrm{Js}}{4 \times 3.14 \times 100 \times 10^{-12} \mathrm{~m}} \\ & \Delta p=0.527 \times 10^{-24} \mathrm{~kg} \mathrm{~m} / \mathrm{s} \end{aligned} $

Among the given options, statement given in I and III are correct, while II is incorrect. The correct form of II is, 3rd orbit radius

$ \begin{aligned} & r_n=n^2 \times r_1 \\ & r_3=9 \times 52.9=476.1 \mathrm{pm} \end{aligned} $

Kinetic energy, when $x$ frequency light is irradiated

$ z=h\left(x-f_0\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Kinetic energy, when $y$ frequency light is irradiated

$ \frac{z}{3}=h\left[y-f_0\right] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

divide Eq. (i) and (ii)

$ \frac{z}{\frac{z}{3}}=\frac{h\left(x-f_0\right)}{h\left(y-f_0\right)} \Rightarrow 3=\frac{x-f_0}{y-f_0} $

On solving for $f_0$

$ f_0=\frac{3 y-x}{2} $

Among the given statements, statements given is II and III are correct. While statement I is incorrect. The correct form of statement I is Isotopes of an element shows same chemical behaviour.

To find the sum of angular and radial nodes for a $4d$ orbital, follow these steps:

Radial Nodes Calculation:

The formula for determining the number of radial nodes is $ n - l - 1 $.

For a $4d$ orbital, the principal quantum number $ n = 4 $ and the azimuthal quantum number $ l = 2 $.

Therefore, the number of radial nodes is:

$ 4 - 2 - 1 = 1 $

Angular Nodes Calculation:

The number of angular nodes is simply given by the azimuthal quantum number $ l $.

Hence, for the $4d$ orbital:

$ l = 2 $

Sum of Angular and Radial Nodes:

Adding the two types of nodes gives:

$ 2 + 1 = 3 $

So, the sum of angular and radial nodes for a $4d$ orbital is 3.

Given:

Position uncertainty: $\Delta x = 0.002\, \text{nm} = 0.002 \times 10^{-9}\, \text{m}$

Planck's constant: $h = 6.626 \times 10^{-34}\, \text{Js}$

According to the Heisenberg uncertainty principle:

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

Solving for the uncertainty in momentum ($\Delta p$):

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

Substitute the given values into the equation:

$ \Delta p = \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 0.002 \times 10^{-9}} $

Calculating the above expression results in:

$ \Delta p = 2.637 \times 10^{-23}\, \text{kg m/s} $

This is the uncertainty in the momentum of the electron.

To determine the de Broglie wavelength of an electron with a kinetic energy of 2.5 eV, we use the following information:

Kinetic Energy (KE) = 2.5 eV

Mass of electron ($M_e$) = $9 \times 10^{-31} \, \mathrm{kg}$

Conversion factor: $1 \mathrm{eV} = 1.6 \times 10^{-19} \, \mathrm{J}$

The de Broglie wavelength ($\lambda$) is given by:

$ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2 M_e \times \mathrm{KE}}} $

where:

$ p = \sqrt{2 M_e \times \mathrm{KE}} $

Therefore, substituting the given values, we find:

$ \lambda = \frac{h}{\sqrt{2 \times 9 \times 10^{-31} \times 2.5 \times 1.6 \times 10^{-19}}} $

Simplifying the expression inside the square root:

$ \lambda = \frac{h}{\sqrt{72} \times 10^{-25}} = \frac{h \times 10^{25}}{\sqrt{72}} \, \mathrm{m} $

The energy of the nth orbit for an atom with atomic number $ Z $ is given by:

$ E_n = -R_{\text{H}} \left(\frac{Z^2}{n^2}\right) $

Here, $ R_{\text{H}} $ is the Rydberg constant.

For the ground state of different ions and atoms, where $ n = 1 $:

Lithium ion ($\mathrm{Li}^{2+}$):

$ Z = 3 $, $ n = 1 $

$ E_{\mathrm{Li}^{2+}} = -R_{\text{H}} \left(\frac{9}{1}\right) $

$ = -9R_{\text{H}} $

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

$ Z = 2 $, $ n = 1 $

$ E_{\mathrm{He}^{+}} = -R_{\text{H}} \left(\frac{4}{1}\right) $

$ = -4R_{\text{H}} $

Hydrogen atom (H):

$ Z = 1 $, $ n = 1 $

$ E_{\mathrm{H}} = -R_{\text{H}} $

The ratio of the ground state energies for $\mathrm{Li}^{2+}$, $\mathrm{He}^{+}$, and $ \mathrm{H} $:

$ E_{\mathrm{Li}^{2+}} : E_{\mathrm{He}^{+}} : E_{\mathrm{H}} = 9 : 4 : 1 $

Given that the longest wavelength of the Paschen series for the $\mathrm{Li}^{2+}$ ion is $x$ Å, we use the formula for the wavelength of spectral lines:

$ \frac{1}{\lambda} = R\left[\frac{1}{n_1^2} - \frac{1}{n_2^2}\right] Z^2 \tag{i} $

Paschen Series for $\mathrm{Li}^{2+}$:

For the longest wavelength in the Paschen series, the transitions are from $n_2 = 4$ to $n_1 = 3$.

The atomic number $Z$ for lithium is 3.

The formula becomes:

$ \frac{1}{x} = R\left[\frac{1}{3^2} - \frac{1}{4^2}\right] \times 3^2 $

Calculating the values:

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

$ \frac{1}{x} = R\left[\frac{7}{144}\right] \times 9 $

Solving for $R$:

$ R = \frac{16}{7x} \, (\text{\AA}^{-1}) $

Lyman Series for Hydrogen:

For the longest wavelength in the Lyman series, the transition is from $n_2 = 2$ to $n_1 = 1$.

The atomic number $Z$ for hydrogen is 1.

Using the value derived for $R$:

$ \frac{1}{\lambda} = \frac{16}{7x}\left[\frac{1}{1} - \frac{1}{4}\right] $

Further simplification:

$ \frac{1}{\lambda} = \frac{16}{7x} \times \frac{3}{4} $

Therefore, the longest wavelength $\lambda$ for the Lyman series in hydrogen is:

$ \lambda = \frac{7x}{12} \, \text{\AA} $

This guides students in calculating the Lyman series' longest wavelength based on known parameters from the Paschen series for the $\mathrm{Li}^{2+}$ ion.

The de Broglie wavelength for an electron is given by the formula:

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

where $ E $ is the kinetic energy of the electron.

According to the photoelectric effect, the kinetic energy $ E $ of the photoelectron can be found by subtracting the work function of the metal from the energy of the incident photon:

$ E = hv - hv_0 $

This implies:

$ E \propto (v - v_0) $

Thus, we have:

$ \lambda \propto \frac{1}{\sqrt{E}} \Rightarrow \lambda \propto \frac{1}{\sqrt{v - v_0}} $

According to Bohr's postulate, the radius is given by $r=0.529 \frac{n^2}{Z} \mathop A\limits^ \circ$ (where, $Z$ is atomic number and $n=$ orbit)

For 4th orbit of $\mathrm{He}^{+}$ion

$ r_4=0.529\left[\frac{16}{2}\right] \mathop A\limits^ \circ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

For 3rd orbit of $\mathrm{He}^{+}$ion

$ r_3=0.529\left[\frac{9}{2}\right] \mathop A\limits^ \circ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Eqs. (i) - (ii)

$ \begin{aligned} & r_4-r_3=\frac{0.529}{2}[16-9] \times 10^{-10} \mathrm{~m} \\ & \Delta r=1.85 \times 10^{-10} \mathrm{~m} \end{aligned} $

In the context of the photoelectric effect, the energy of an incident photon is used to overcome the work function of the metal and to provide kinetic energy (KE) to the ejected photoelectrons. This relationship is described by Einstein's photoelectric equation:

$ h\nu = h\nu_0 + \text{KE} $

Here, $ \nu $ is the frequency of the incident light, $\nu_0$ is the threshold frequency, and $ h $ is Planck's constant.

The kinetic energy (KE) of the ejected photoelectrons can be expressed as:

$ \text{KE} = \frac{1}{2}mv^2 = h\nu - h\nu_0 $

Since frequency $ \nu $ can be related to wavelength $\lambda$ by the equation $\nu = \frac{c}{\lambda}$, where $ c $ is the speed of light, the equation becomes:

$ \frac{1}{2}mv^2 = hc\left(\frac{1}{\lambda} - \frac{1}{\lambda_0}\right) $

Solving for $ v $, the velocity of the ejected photoelectrons, we have:

$ v = \sqrt{\frac{2hc}{m} \left(\frac{\lambda_0 - \lambda}{\lambda \lambda_0}\right)} $

Thus, the velocity of the ejected photoelectrons is derived from this equation, showing the relationship between the incident light wavelength, the threshold wavelength, and the resulting kinetic energy imparted to the electrons.

The energy of the nth orbital of an atom, as described by Bohr's postulate, is given by the following formula:

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

In this equation:

$ Z $ is the atomic number, which for lithium (Li) is 3.

$ n $ is the principal quantum number, and for the third orbit, $ n = 3 $.

Substituting these values into the formula:

$ E_n = \frac{-2.18 \times 10^{-18} \times 3^2}{3^2} \, \text{J} $

$ E_n = \frac{-2.18 \times 10^{-18} \times 9}{9} \, \text{J} $

$ E_n = -2.18 \times 10^{-18} \, \text{J} $

Thus, the energy of the third orbit of the $\text{Li}^{2+}$ ion is $-2.18 \times 10^{-18} \, \text{J}$.

$ \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 $(5, 0, 0, \frac{1}{2})$.

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 $=11$

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