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 $\quad x=4$

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

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 maximum number of possible electrons in a subshell with $n=3, l=2$ are 10 .

Because for, $n=3, l=2$, i.e. $d$-subshell has 10 electrons and 5 -orbital.

Hence, option (a) is the correct answer.

Given, uncertainty in position $(\Delta x)=1 \times 10^4 \mathrm{~m}$

uncertainty in velocity

$ (\Delta v)=0.627 \times 10^{-20} \mathrm{~ms}^{-1} $

Also,

$ \begin{aligned} & \because \quad \Delta x \cdot \Delta v=\frac{h}{4 \pi \cdot m} \\ & \text { } \begin{aligned} where, h & =\text { Plank's constant } \\ & =6.627 \times 10^{-34} \mathrm{Js} \\ m & =\text { to find (mass of particle) } \\ \therefore \quad m & =\frac{h}{\Delta x \cdot \Delta v \cdot 4 \pi} \\ & =\frac{6.627 \times 10^{-34}}{10^{-8} \times 6.627 \times 10^{-20} \times 4 \pi} \\ & =\frac{10^{-34} \times 10^{28}}{4 \pi} \text { in } \mathrm{kg} \\ \therefore \quad m & =\frac{10^{-34} \times 10^{28}}{4 \pi} \Rightarrow m=\frac{10^{-6}}{4 \pi} \mathrm{~kg} \end{aligned} \end{aligned} $

Hence, option (d) is the correct answer.

From de-Broglie's equation, $\lambda=\frac{h}{m v}=\frac{h}{\sqrt{2 m \times \mathrm{KE}}}$

⇒ For an electrons, $\lambda \propto \frac{1}{\sqrt{\mathrm{KE}}} \Rightarrow \frac{\lambda_{16}}{\lambda_9}=\sqrt{\frac{9}{16}}=3: 4$