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.

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

$E_n = -\frac{R_{\mathrm{H}}}{n^2}$

We are told that the energy of the atom is equal to $-\frac{R_{\mathrm{H}}}{16}$. To find the principal quantum number $ n $, we compare the two expressions:

$-\frac{R_{\mathrm{H}}}{n^2} = -\frac{R_{\mathrm{H}}}{16}$

Canceling out $ -R_{\mathrm{H}} $ from both sides gives:

$\frac{1}{n^2} = \frac{1}{16}$

Therefore, $ n^2 = 16 \Rightarrow n = 4 $.

In a hydrogen atom, all orbitals that share the same principal quantum number $ n $ have the same energy. This means they are called degenerate. The total number of these degenerate orbitals is given by $ n^2 $.

So, for $ n = 4 $:

Degeneracy $ = n^2 = 4^2 = 16 $.

The de-Broglie wavelength is given by the formula:

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

According to the question, the wavelength is also given as:

$\lambda = 10 \sqrt{\frac{h}{m}}$

Now, equating the two expressions for $ \lambda $:

$\frac{h}{mV} = 10 \sqrt{\frac{h}{m}}$

Multiply both sides by $ mV $:

$h = 10 \, mV \sqrt{\frac{h}{m}}$

Divide both sides by 10:

$\frac{h}{10} = mV \sqrt{\frac{h}{m}}$

Square both sides to remove the square root:

$\left(\frac{h}{10}\right)^2 = (mV)^2 \times \frac{h}{m}$

Simplify the equation:

$\frac{h^2}{100} = m^2V^2 \frac{h}{m}$

$\frac{h}{100} = mV^2$

Now, solve for $ V $:

$V = \sqrt{\frac{h}{100m}}$

To find the ratio of speed to mass, $ \frac{V}{m} $:

$\frac{V}{m} = \frac{1}{m} \sqrt{\frac{h}{100m}} = \frac{\sqrt{h}}{10 \, m^{3/2}}$

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} $