Atomic Structure

Statement I is correct, but statement II is not correct. The correct form of statement II is, The wavelength of X-ray is smaller than the wavelength of microwaves.

Let principal quantum number be $n_1$ and $n_2$.

For $r_1$

$ \begin{aligned} \Rightarrow \quad n_1^2 & =\frac{r_1}{a_0}=\frac{1.3225 \times 10^{-9}}{0.529 \times 10^{-10}}=25 \\ n_1 & =5 \end{aligned} $

For $r_2$

$ \begin{aligned} \Rightarrow \quad n_2^2 & =\frac{r_2}{a_0}=\frac{0.2116 \times 10^{-9}}{0.529 \times 10^{-10}}=4 \\ n_2 & =\sqrt{4}=2 \end{aligned} $

For $n_1, E_1=-\frac{13.6}{n_1^2} \mathrm{eV}=\frac{-13.6}{(5)^2}=-0.544 \mathrm{eV}$

For $n_2, E_2=\frac{-13.6}{n_2^2}=\frac{-13.6}{4}=-3.4 \mathrm{eV}$

Energy of emitted radiation,

$ \begin{aligned} & E=E_1-E_2=2856 \times 1.602 \times 10^{-19} \\ & E \approx 0.4578 \times 10^{-18} \mathrm{~J} \end{aligned} $

Formula for orbit radius:

The radius of the $n$th orbit in a hydrogen-like ion is:

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

Here, $r_n$ is the radius of the orbit, $n$ is the orbit number, $a_0$ is the Bohr radius, and $Z$ is the atomic number.

Radius for Hydrogen atom ($Z=1$):

For the second orbit of hydrogen ($n = 2$):

$ r_{\mathrm{H}} = \frac{2^2 a_0}{1} = 4 a_0 $

Radius for other ion:

For an ion $x$ with atomic number $Z$ and orbit number $n$:

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

Set radii equal to each other:

Since the radii are equal:

$ \frac{n^2 a_0}{Z} = 4 a_0 $

Solve for $n^2$:

Divide both sides by $a_0$:

$ \frac{n^2}{Z} = 4 $

Multiply both sides by $Z$:

$ n^2 = 4Z $

Find values for $n$ and $Z$:

Try $Z = 4$ (which is for Be$^{3+}$):

$ n^2 = 4 \times 4 = 16 $

So, $n = 4$.

Answer:

The orbit number $n$ is 4, and the ion $x$ is $\mathrm{Be}^{3+}$.

According to the question, given values are

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

$ N_A=6 \times 10^{23} \mathrm{~mol}^{-1} $

Wavelength,

$ \lambda=331.5 \mathrm{~nm}=331.5 \times 10^{-9} \mathrm{~m} $

Kinetic energy (per mole)

$ =1.2 \times 10^5 \mathrm{~J} / \mathrm{mol} $

The energy of a single photon is given by the formula, $E=\frac{h c}{\lambda}$

Substituting the given values,

$ \begin{aligned} E & =\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{331.5 \times 10^{-9}} \\ & =6.00 \times 10^{-19} \mathrm{~J} \end{aligned} $

Energy per mole of photons,

$ \begin{aligned} E_{\mathrm{mol}} & =6.00 \times 10^{-19} \times 6 \times 10^{23} \\ & =3.6 \times 10^5 \mathrm{~J} / \mathrm{mol} \end{aligned} $

Using photoelectric equation (per mole)

Work function $(\mathrm{mol})=E_{\text {mol }}-\mathrm{KE}$

$ \begin{aligned} & =3.6 \times 10^{-5}-1.2 \times 10^5 \\ & =2.4 \times 10^5 \mathrm{~J} / \mathrm{mol} \end{aligned} $

Since, $1 \mathrm{eV}=96,485 \mathrm{~J} / \mathrm{mol}$

$ \phi=\frac{2.4 \times 10^5}{96485} \approx 2.5 \mathrm{eV} $

In the atomic spectrum of hydrogen, the wavenumber corresponding to an electronic transition is given by the Rydberg formula.

$ v=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) $

For transition (i) : $n=4 \rightarrow n=2$ the corresponding wavelength is $\lambda_1$.

For transition (ii) : $n=3 \rightarrow n=1$, the corresponding wavelength is $\lambda_2$.

Using the Rydberg formula

$ \begin{aligned} & \frac{1}{\lambda_1}=R_H\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=R_H\left(\frac{1}{4}-\frac{1}{16}\right) \\ & =R_H\left(\frac{3}{16}\right) \\ & \frac{1}{\lambda_2}=R_H\left(\frac{1}{1^2}-\frac{1}{3^2}\right)=R_H\left(1-\frac{1}{9}\right)=R_H\left(\frac{8}{9}\right) \end{aligned} $

We now calculate the difference in wavelengths

$ \begin{aligned} \lambda_1-\lambda_2 & =\frac{1}{v_1}-\frac{1}{v_2} \\ & =\frac{1}{\left(\frac{3 R_H}{16}\right)}-\frac{1}{\left(\frac{8 R_H}{9}\right)}=\frac{16}{3 R_H}-\frac{9}{8 R_H} \end{aligned} $

Taking LCM,

$ \begin{aligned} \lambda_1-\lambda_2 & =\frac{(16.8-9.3)}{24 R_H} \\ & =\frac{(128-27)}{24 R_H}=\frac{101}{24 R_H} \end{aligned} $

Thus, $\lambda_1-\lambda_2=\frac{1}{R_H}\left(\frac{101}{24}\right)$

Therefore, the correct option is (c).

Energy of photon

$ E=\frac{1240}{\lambda(\mathrm{in} \mathrm{~nm})}=\frac{1240}{310}=4.0 \mathrm{eV} $

Given work functions

• $M_1=4.8 \mathrm{eV}$

• $M_2=4.3 \mathrm{eV}$

• $M_3=4.75 \mathrm{eV}$

• $\mathrm{M}_4=3.75 \mathrm{eV}$

Only metals with $\phi \leq 4.0 \mathrm{eV}$ will emit electrons.

$M_1, M_2$ and $M_3$ have work functions greater than 4.0 eV thus, do not show no photoelectric effect.

Energy in the $n$th orbit is given by

$ E_n=\frac{E_1}{n^2} $

For $n=5$,

$ \begin{aligned} E_5 & =\frac{-2.18 \times 10^{-18} \mathrm{~J}}{\left(S^2\right.} \\ & =-8.72 \times 10^{-20} \mathrm{~J} \\ \Delta E & =E_5-E_1 \\ & =\left(-8.72 \times 10^{-20}\right)-\left(-2.18 \times 10^{-18}\right) \\ & =2.0928 \times 10^{-18} \mathrm{~J} \\ v & =\frac{\Delta E}{n}=\frac{2.0928 \times 10^{-18}}{6.6 \times 10^{-34}} \\ & =3.17 \times 10^{15} \mathrm{~Hz} \end{aligned} $

Step 1: Write the electron configuration of Strontium ($\mathrm{Sr}$, $Z=38$)

$ \begin{aligned} \mathrm{Sr}(Z=38) = 1s^2 \; 2s^2 \; 2p^6 \; 3s^2 \; 3p^6 \; 4s^2 \; 3d^{10} \; 4p^6 \; 5s^2 \end{aligned} $

Step 2: Count electrons with $l=0$

Electrons with $l=0$ are in $s$ orbitals. Add up all the electrons in $s$ orbitals: $ x = 2 (1s) + 2 (2s) + 2 (3s) + 2 (4s) + 2 (5s) = 10 $ So, $x=10$

Step 3: Count electrons with $l=2$

Electrons with $l=2$ are in $d$ orbitals. Look for the $d$ orbital electrons: $ y = 10 \text{ (from $3d^{10}$)} $ So, $y=10$

Step 4: Find $x-y$

Subtract to get: $ x-y = 10-10 = 0 $

Step 5: Find $y \cdot (x-y)$

Multiply $y$ and $x-y$: $ y \cdot (x-y) = 10 \times 0 = 0 $

For hydrogen atom, radius is given by

$ \begin{aligned} & r_n=n^2 \cdot a_0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & a_0=52.9 \mathrm{pm} \text { (Bohr's radius) } \\ & r_n=476.1 \mathrm{pm} \end{aligned} $

Substituting the given values in Eq. (i),

$ n^2 \approx 9, n=3 $

Transition is to $n=3$, so it is a Paschen series.

Among the given statements, statement given in option

(b) is incorrect. It's correct form is,

The probability density of electron is expressed by $|\psi|^2$.

Radius of $n$th orbit of hydrogen and hydrogen like atom, $r_n=\frac{n^2 a_0}{Z}$

For $n=2, r_n=x \mathrm{pm}, Z=1 \quad$ (for H-atom)

$ x=\frac{2^2\left(a_0\right)}{1} \Rightarrow a_0=\frac{x}{4} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

For $n=3, Z=2, r_n=$ ?

(for $\mathrm{He}^{+}$atom)

$ \begin{aligned} r_n & =\frac{(3)^2 a_0}{2}=\frac{9 a_0}{2} \\ \Rightarrow \quad r_n & =\frac{9 \times x}{2 \times 4}=\frac{9 x}{8} \end{aligned} $

Given, $\lambda=310 \mathrm{~nm}$

$ \begin{aligned} & =310 \times 10^{-9} \mathrm{~m} \\ & \phi=3.55 \mathrm{eV} \Rightarrow v=x \times 10^5 \mathrm{~m} / \mathrm{s} \\ & x=? \end{aligned} $

Energy of incident photon,

$ \begin{aligned} & E=\frac{h c}{\lambda}=\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{310 \times 10^{-9}} \\ & E=6.412 \times 10^{-19} \mathrm{~J} \end{aligned} $

Kinetic energy of photoelectrons,

$ \mathrm{KE}=E-\phi $

$ \begin{aligned} & \mathrm{KE}=6.412 \times 10^{-19}-3.55 \times 1.6 \times 10^{-19} \\ & \mathrm{KE}=0.732 \times 10^{-19} \mathrm{~J} \end{aligned} $

Velocity of photoelectrons,

$ \begin{aligned} v & =\sqrt{\frac{2 \mathrm{KE}}{m_e}} \\ & =\sqrt{\frac{2 \times 0.732 \times 10^{-19}}{9 \times 10^{-31}}} \\ v & =4 \times 10^5 \mathrm{~m} / \mathrm{s} \end{aligned} $

or $\quad x=4$

$ \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.

Angular nodes are given by ' $ l $ ' quantum number and radial nodes are determined by $ n-l-1 $, where $ n $ is principal quantum number, $ l $ is azimuthal quantum number.

Given that the radius of the first orbit of a hydrogen-like ion is $1.763 \times 10^{-2} \, \text{nm}$, we can find the energy associated with that orbit using Bohr's postulates.

Calculate Atomic Number $Z$:

According to Bohr's model, the radius of the $n$-th orbit is given by:

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

For the first orbit ($n = 1$), we have:

$ r_1 = \frac{a_0}{Z} $

Given $r_1 = 1.763 \times 10^{-2} \, \text{nm}$ and the Bohr radius $a_0 = 0.0529 \, \text{nm}$, we can solve for $Z$:

$ Z = \frac{a_0}{r_1} = \frac{0.0529 \, \text{nm}}{1.763 \times 10^{-2} \, \text{nm}} \approx 3 $

Calculate the Energy $E_1$:

Using Bohr's expression for the energy of the $n$-th orbit:

$ E_n = \frac{Z^2 \times (-13.6 \, \text{eV})}{n^2} $

For the ground state ($n = 1$):

$ E_1 = \frac{3^2 \times (-13.6 \, \text{eV})}{1^2} = -122.4 \, \text{eV} $

Convert Energy to Joules:

We convert the energy from electron volts to joules using the conversion factor $1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$:

$ E_1 = -122.4 \times 1.6 \times 10^{-19} = -1.962 \times 10^{-17} \, \text{J} $

Thus, the energy associated with the first orbit is $-1.962 \times 10^{-17} \, \text{J}$.

To find the momentum of an electron given its wavelength, we can use the de Broglie relation.

Given:

Wavelength, $ \lambda = 10^{3} \, \text{nm} $, which can be converted to meters as follows:

$ \lambda = 10^{3} \times 10^{-9} = 10^{-6} \, \text{m} $

Planck's constant, $ h = 6.625 \times 10^{-34} \, \text{Js} $

The de Broglie wavelength formula is:

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

Solving for momentum $ p $, we rearrange the formula:

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

Substituting the given values into the equation:

$ p = \frac{6.625 \times 10^{-34}}{10^{-6}} $

Calculating the result:

$ p = 6.625 \times 10^{-28} \, \text{kg m/s} $

Therefore, the momentum of the electron is $ 6.625 \times 10^{-28} \, \text{kg m/s} $.

Statement $ I $ is correct but Statement II is incorrect. For single electron species energy depends only on value of $ n $ and not $ l $.

But in case He , the energy depends on both $ n $ and $ l $ for $ 2 s $ energy is $ 2+0=2 $ and for $ 2 p, 2+1=3 $.

Hence, $ 2 p $ has more energy.

The wave number is given by the formula,

$ \overline{\mathrm{v}}=\frac{1}{\lambda}=R Z^{2}\left[\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right] $

Case (i) First spectral line of Lyman series of $ \mathrm{He}^{+} $has wave number $ x \mathrm{~m}^{-1} $.

Substitute the value in equation (i).

$ x=R(2)^{2}\left[\frac{1}{(1)^{2}}-\frac{1}{(2)^{2}}\right] $

As $ n_{i}=2 $ and $ n_{f}=1 $ for first spectral line of Lyman series

$ x=R \times 4 \times \frac{3}{4} \Rightarrow x=3 R $

Case (ii) 2nd spectral line of Balmer series of $ \mathrm{Li}^{2+} $, let wave number be $ y $.

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

As $ n_{i}=4 $ and $ n_{f}=2 $ for 2 nd spectral line of Balmer series for $ \mathrm{Li}^{2+} $ ion.

Substitute the value of $ R $ from Eq. (ii) in Eq. (iv).

$ y=\frac{x}{3} \times(3)^{2} \times \frac{3}{16} \Rightarrow y=\frac{9 x}{16} \mathrm{~m}^{-1} $

Given,

$ \begin{array}{l} \Delta x=10^{-33} \mathrm{~m}, m=10 \mathrm{~g}, v=52.5 \mathrm{~m} / \mathrm{s}, \\\\ h=6.6 \times 10^{-34} \mathrm{Js} . \end{array} $

According to Heisenberg's uncertainty principle,

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

lst step is to calculate uncertainty in velocity

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

Substitute all values in Eq. (ii),

$ \begin{aligned} \Delta v & =\frac{6.6 \times 10^{-34}}{4 \pi \times 10 \times 10^{-3} \times 10^{-33}} \\\\ & =5.25 \mathrm{~m} / \mathrm{s} \end{aligned} $

$ \% $ of accuracy is

$ \frac{\Delta v}{v} \times 100 \Rightarrow \frac{5.25}{52.5} \times 100=10 \% $

Given,

Radiation frequency,

$ \mathrm{v}=1.0 \times 10^{15} \mathrm{~Hz} $

Kinetic energy, $\mathrm{KE}=2 \times 10^{-19} \mathrm{~J}$

According to Einstein's equation for photoelectric effect,

$ \begin{array}{ll} h v=h v_0+\frac{1}{2} m v^2 \\\\ \text { or } h v=h v_0+\mathrm{KE} \quad \ldots \text { (i) } \end{array} $

where, $v_0$ is threshold frequency

$ \frac{h v-\mathrm{KE}}{h}=v_0 \quad \ldots \text { (ii) } $

Substitute all the values, in above formula,

$ \begin{aligned} & \frac{6.6 \times 10^{-34} \times 1 \times 10^{15}-2 \times 10^{-19}}{6.6 \times 10^{-34}}=v_0 \\\\ & v_0=6.97 \times 10^{14} \mathrm{~Hz} \end{aligned} $

Energy according to Bohr's postulate is given by

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

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

$ E=-13.6 \frac{(3)^2}{(3)^2} ; E=-13.6 \mathrm{eV} $

For $\mathrm{Be}^{3+},(n=4, Z=4)$

$ E=-13.6 \frac{(4)^2}{(4)^2} ; E=-13.6 \mathrm{eV} $

Hence, $\mathrm{Li}^{2+}$ and $\mathrm{Be}^{3+}$ have same energy.

In the Balmer series of the hydrogen atom, the electron transition occurs from a higher energy level ($n \geq 3$) to the second energy level ($n = 2$). The wavelength ($\lambda$) for these transitions is given by the Rydberg formula:

$ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right) $

For different $n$ values, this can be used to calculate the wavelength:

For $n = 3$

$ \frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9 - 4}{36} \right) = \frac{5R}{36} $

Therefore, $ \lambda = \frac{36}{5R} $

For $n = 4$

$ \frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{16 - 4}{64} \right) = \frac{3R}{16} $

Therefore, $ \lambda = \frac{16}{3R} $

For $n = 5$

$ \frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{25} \right) = R \left( \frac{25 - 4}{100} \right) = \frac{21R}{100} $

Therefore, $ \lambda = \frac{100}{21R} $

For $n = 6$

$ \frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{36} \right) = R \left( \frac{36 - 4}{144} \right) = \frac{32R}{144} = \frac{8R}{36} $

Therefore, $ \lambda = \frac{36}{8R} = \frac{9}{2R} $

Clearly, the correct option, based on our calculations with the transitions that correspond to the Balmer series, is from the option for $n = 5$:

Option B: $\frac{100}{21R}$

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