Structure of Atom

$ \lambda_{\mathrm{d}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK.E.}}} $

Here, both particles are accelerated through the same potential difference of 200 keV, so they gain the same kinetic energy (K.E.).

So, in the formula, $\mathrm{h}$ and K.E. are the same for both particles. Only the mass $\mathrm{m}$ is different.

Therefore, $ \lambda_{\mathrm{d}} \propto \frac{1}{\sqrt{\mathrm{~m}}} $

Now, $ \begin{aligned} & \frac{\left(\lambda_{\mathrm{d}}\right)_{\mathrm{m}_1}}{\left(\lambda_{\mathrm{d}}\right)_{\mathrm{m}_2}}=\sqrt{\frac{\mathrm{m}_2}{\mathrm{~m}_1}}=\sqrt{4}=2 \\ & \left(\lambda_{\mathrm{d}}\right)_{\mathrm{m}_1}=2\left(\lambda_{\mathrm{d}}\right)_{\mathrm{m}_2} \end{aligned} $

So $\mathrm{x}=2$.

For hydrogen, the energy difference for a spectral line is given by

$\Delta E=13.6 \times Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

Step 1: Identify the transitions

In the Lyman series, electrons fall to $n_1=1$. The first line $L_1$ is the transition $2 \to 1$.

In the Balmer series, electrons fall to $n_1=2$. The first line $B_1$ is the transition $3 \to 2$.

Step 2: Find energy of $L_1$

$\begin{aligned} & \Delta \mathrm{E}\left(\mathrm{L}_1\right)=13.6 \times \mathrm{Z}^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=13.6 \mathrm{Z}^2 \times \frac{3}{4} \end{aligned}$

Step 3: Find energy of $B_1$

$\begin{aligned} & \Delta \mathrm{E}\left(\mathrm{B}_1\right)=13.6 \times \mathrm{Z}^2 \times\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=13.6 \times \mathrm{Z}^2 \times \frac{5}{4 \times 9} \end{aligned}$

Step 4: Take the ratio to get $x$

The factor $13.6Z^2$ cancels in the ratio.

$\begin{aligned} & \frac{\Delta \mathrm{E}\left(\mathrm{L}_1\right)}{\Delta \mathrm{E}\left(\mathrm{B}_1\right)}=\frac{3}{5} \times 9=\frac{27}{5}=\mathrm{x} \end{aligned}$

Step 5: Write in the asked form

$\begin{aligned} & \mathrm{x}=\left(\frac{27}{5} \times 10\right) \times 10^{-1}=54 \times 10^{-1} \end{aligned}$

The formula for the wavelength of light absorbed in a transition for a hydrogen-like species is:

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

For $ X^{a+} $:

$\frac{1}{\lambda} = R_H (z_x)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \qquad \ldots\ldots\ldots (1)$

For $ Y^{b+} $:

Since wavelength is $9\lambda$,

$\frac{1}{9\lambda} = R_H (z_y)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \qquad \ldots\ldots\ldots (2)$

Dividing Equation (1) by Equation (2):

$9 = \frac{z_x^2}{z_y^2} \times \frac{\left( \frac{3}{4} \right)}{\left( \frac{3}{16} \right)}$

$9 = \frac{z_x^2}{z_y^2} \times 4$

$\frac{z_x^2}{z_y^2} = \frac{9}{4}$

$\frac{z_x}{z_y} = \frac{3}{2}$

Choosing the smallest integer values that satisfy the ratio:

Let $ z_x = 3 $ and $ z_y = 2 $.

For hydrogen-like ions, $ z = \text{atomic number} = a + 1 $ (since one electron is left).

Thus, for $ X^{a+} $: $ a = 3 - 1 = 2 $

And for $ Y^{b+} $: $ b = 2 - 1 = 1 $

Therefore, the lowest possible value of $ a + b = 2 + 1 = 3 $.

To find the energy of an electron in the first excited state of a Be³⁺ ion, we can use the formula for the energy of an electron in a hydrogen-like atom:

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

Given:

Energy of the first orbit (ground state) of the hydrogen atom: $ \mathrm{E}_1 = -13.6 \text{ eV} $ (where $ Z = 1 $ and $ n = 1 $).

For Be³⁺:

Atomic number $ Z = 4 $.

First excited state corresponds to $ n = 2 $.

Calculation for Be³⁺:

The energy ratio between the hydrogen atom and the Be³⁺ ion can be written as:

$ \frac{\mathrm{E}_{\mathrm{H}}}{\mathrm{E}_{\mathrm{Be}^{+3}}} = \frac{Z_1^2}{n_1^2} \times \frac{n_2^2}{Z_2^2} $

Substitute the known values:

$ \frac{-13.6}{\mathrm{E}_{\mathrm{Be}^{+3}}} = \frac{1^2}{1^2} \times \frac{2^2}{4^2} $

$ \frac{-13.6}{\mathrm{E}_{\mathrm{Be}^{+3}}} = \frac{1}{1} \times \frac{4}{16} $

Solving this gives:

$ \mathrm{E}_{\mathrm{Be}^{+3}} = -13.6 \times 4 = -54.4 \text{ eV} $

The magnitude of the energy of the electron in the first excited state of Be³⁺ is $\left| -54.4 \right| = 54.4 \text{ eV}$, which is approximately 54 eV when rounded to the nearest integer.

To find the uncertainty in the velocity of the electron within an atomic nucleus of diameter $10^{-15} \ \text{m}$, we can use Heisenberg's uncertainty principle. The principle is expressed as:

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

Here, $\Delta x$ is the uncertainty in position, $m$ is the mass of the electron, $\Delta v$ is the uncertainty in velocity, and $h$ is Planck's constant.

Given:

• $ \Delta x = 10^{-15} \ \text{m} $


• $ m = 9.1 \times 10^{-31} \ \text{kg} $


• $ h = 6.626 \times 10^{-34} \ \text{Js} $


• $ \pi = 3.14 $

We need to find $\Delta v$. Rearrange the uncertainty principle to solve for $\Delta v$:

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

Plug in the numbers:

$ \Delta v = \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 10^{-15}} $

First, calculate the denominator:

$ 4 \times 3.14 \times 9.1 \times 10^{-31} \times 10^{-15} = 114.392 \times 10^{-46} $

Now, perform the division:

$ \Delta v = \frac{6.626 \times 10^{-34}}{114.392 \times 10^{-46}} = \frac{6.626}{114.392} \times 10^{12} $

Perform the division:

$ \Delta v \approx 0.0579 \times 10^{12} \ \text{m/s} = 57.97 \times 10^9 \ \text{m/s} $

So, the uncertainty in the velocity of the electron is:

$ 58 \times 10^{9} \ \text{m/s} \quad \text{(nearest integer)} $

The wavenumber of a radiation is defined as the number of wavelengths per unit distance and is the reciprocal of the wavelength. Wavenumber is commonly represented in units of $\mathrm{cm}^{-1}$.

First, convert the given wavelength from angstroms ($\mathop A\limits^o $) to centimeters (cm). We know that:

$1 \mathop A\limits^o = 10^{-8} \, \text{cm}$

Given wavelength is 5800 $\mathop A\limits^o $:

$5800 \mathop A\limits^o = 5800 \times 10^{-8} \, \text{cm}$

Now calculate the wavenumber ($\tilde{\nu}$) which is the reciprocal of the wavelength:

$\tilde{\nu} = \frac{1}{{5800 \times 10^{-8} \, \text{cm}}}$

Simplify the expression:

$\tilde{\nu} = \frac{1}{5800 \times 10^{-8} \, \text{cm}} = \frac{10^8}{5800} \, \mathrm{cm}^{-1}$

Now, divide the numerator by the denominator to calculate the precise value:

$\tilde{\nu} = \frac{10^8}{5800} \approx 1.724 \times 10^4 \, \mathrm{cm}^{-1}$

Here, it is given that the wavenumber is in the form of $x \times 10 \, \mathrm{cm}^{-1}$, so $x$ would be the value we calculated divided by 10:

$x = \frac{1.724 \times 10^4}{10} = 1724$

Thus, the integer answer for the value of $x$ is:

1724

$\begin{aligned} \lambda & =1.5 \times 4=6 \mathrm{~pm} \\ \nu & =\frac{C}{\lambda} \\ \nu & =\frac{3 \times 10^8}{6 \times 10^{-12}} \\ & =0.5 \times 10^{20} \\ \nu & =5 \times 10^{19} \\ x & =5 \end{aligned}$

To determine the kinetic energy (K.E.) of an electron in the first excited state of a hydrogen atom, we need to understand the relationship between the total energy, potential energy, and kinetic energy in an atom.

In a hydrogen atom, the total energy (E) of an electron in the nth state is given by:

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

For the first excited state, $ n = 2 $. So, plugging in the value:

$E_2 = - \frac{13.6}{2^2} = - \frac{13.6}{4} = - 3.4 \mathrm{~eV}$

This value represents the total energy (E) of the electron in the first excited state. According to the virial theorem for an electron in a Coulomb potential (as in a hydrogen atom), the kinetic energy (K.E.) is equal to the negative of the total energy:

$\mathrm{K.E.} = - E$

Substituting the total energy we calculated:

$\mathrm{K.E.} = - (-3.4) = 3.4 \mathrm{~eV}$

Now, we need to find the value of $x$ in the form of $x \times 10^{-1} \mathrm{~eV}$.

$3.4 \mathrm{~eV} = 34 \times 10^{-1} \mathrm{~eV}$

Therefore, the value of $x$ is 34.

$\begin{aligned} & \lambda=\frac{h}{m v} \\ & \lambda \cdot v=\frac{h}{m} \\ & \frac{\lambda \cdot v^2}{v}=\frac{h}{m} \\ & \frac{v^2}{\text { Frequency }}=\frac{h}{m} \end{aligned}$

$\begin{aligned} & \text { Frequency }=\frac{\mathrm{mv}^2}{\mathrm{~h}}=\frac{2 \mathrm{R}_{\mathrm{H}}}{\mathrm{h}} \\ & =\frac{2 \times 2.18 \times 10^{-18}}{6.6 \times 10^{-34}} \\ & =660.6 \times 10^{13} \mathrm{~Hz} \end{aligned}$

Nearest integer $=661$

If we take $h=6.626 \times 10^{-34}$

Then

$\begin{aligned} \text { Frequency } & =\frac{2 R_H}{h} \\ & =658.01 \times 10^{13} \mathrm{~Hz} \end{aligned}$

But if value of $\mathrm{h}$ is given as $6.6 \times 10^{-34}$ correct answer will be 661.

To find the total number of electrons with the given quantum numbers, let's first understand what each quantum number represents:

• n (Principal Quantum Number): This specifies the energy level or shell of the electron in the atom. Here, $n = 4$ means we are dealing with the fourth energy level.
• l (Azimuthal Quantum Number or Orbital Angular Momentum Quantum Number): This defines the shape of the orbital and is dependent on the value of n. It can take values from 0 to $n-1$. For $n = 4$, $l$ can be 0 (s orbital), 1 (p orbital), 2 (d orbital), or 3 (f orbital).
• $m_l$ (Magnetic Quantum Number): This describes the orientation of the orbital in space and can take values from $-l$ to $+l$, including zero. The condition $\left|\mathrm{~m}_l\right|=1$ implies $m_l = -1$ or $+1$.
• $m_s$ (Spin Quantum Number): This indicates the spin direction of the electron, with possible values of $+\frac{1}{2}$ or $-\frac{1}{2}$. Here, $\mathrm{m}_{\mathrm{s}}=-\frac{1}{2}$ specifies electrons with a downward spin.

Given $\left|\mathrm{~m}_l\right|=1$, this directly implies that we are not considering s orbitals (which have $l = 0$ and hence $m_l = 0$). We are considering p, d, and f orbitals which can have $m_l = 1$ or $m_l = -1$.

For $l = 1$ (p orbitals):

• This corresponds to p orbitals, where $\left|\mathrm{~m}_l\right|=1$ can be achieved with $m_l = -1$ or $m_l = +1$.
• With $\mathrm{m}_{\mathrm{s}}=-\frac{1}{2}$, we only take half of the available m_l values since each orbital can host two electrons with different spin quantum numbers.
• So, we get 2 electrons for p orbitals with the specified conditions.

For $l = 2$ (d orbitals):

• This corresponds to d orbitals, where again $\left|\mathrm{~m}_l\right|=1$ indicates $m_l = -1$ or $m_l = +1$.
• Similarly, we consider only electrons with a spin quantum number of $-\frac{1}{2}$, leading us to again have 2 electrons.

For $l = 3$ (f orbitals):

• f orbitals also can have $m_l = -1$ or $m_l = +1$ satisfying the $\left|\mathrm{~m}_l\right|=1$.
• With $\mathrm{m}_{\mathrm{s}}=-\frac{1}{2}$, like before, this limits us to 2 electrons.

Adding up all the electrons from the p, d, and f orbitals that meet the given criteria, we get:

• $2\ (\mathrm{p\ orbitals}) + 2\ (\mathrm{d\ orbitals}) + 2\ (\mathrm{f\ orbitals}) = 6\ \mathrm{electrons}$

Therefore, the total number of electrons having quantum numbers with $\mathrm{n}=4, \left|\mathrm{~m}_l\right|=1$ and $\mathrm{m}_{\mathrm{s}}=-\frac{1}{2}$ is 6.

$\begin{aligned} & \text { K.E. }=R_H \cdot \frac{z^2}{n^2}=\frac{1}{2} m v^2 \\ & v^2=\frac{2 \times 2.18 \times 10^{-18}}{9.1 \times 10^{-31}} \times \frac{1}{1}=0.479 \times 10^{13} \\ & v=21.88 \times 10^5 \mathrm{~m} / \mathrm{s} \end{aligned}$

Possible combination of first three quantum numbers are

$\begin{aligned} & n=4, l=3, m=0 \\ & n=4, l=2, m=0 \\ & n=4, l=1, m=0 \\ & n=4, l=0, m=0 \end{aligned}$

The de-Broglie wavelength of an electron can be expressed through its relationship with the principal quantum number $n$ in a Bohr orbit. According to de Broglie, the wavelength $ \lambda $ of an electron moving in an orbit can be related to its momentum. However, when considering an electron in an atom, specifically within the Bohr model, we can relate the de-Broglie wavelength to the circumference of the orbit it moves in. The formula representing the de Broglie wavelength of an electron in the $n^{\text{th}}$ orbit of a hydrogen-like atom is:

$ \lambda_n = \frac{2 \pi r_n}{n} $

where

• $ \lambda_n $ is the de-Broglie wavelength of the electron in the $n^{\text{th}}$ orbit,
• $ r_n $ is the radius of the $n^{\text{th}}$ orbit,
• $ n $ is the principal quantum number (which is 4 in this case).

According to Bohr's theory, the radius of the $n^{\text{th}}$ orbit is given by:

$ r_n = n^2 a_0 $

where $ a_0 $ is the Bohr radius.

For the $4^{\text{th}}$ orbit, $n = 4$, so the radius is:

$ r_4 = 4^2 a_0 = 16 a_0 $

Substituting $ r_4 $ into the de-Broglie wavelength formula, we get:

$ \lambda_4 = \frac{2 \pi (16 a_0)}{4} = 8 \pi a_0 $

$\begin{aligned} & \mathrm{E}=\frac{1240}{\lambda(\mathrm{nm})} \mathrm{eV} \\ & =\frac{1240}{242} \mathrm{eV} \\ & =5.12 \mathrm{eV} \\ & =5.12 \times 1.6 \times 10^{-19} \\ & =8.198 \times 10^{-19} \mathrm{~J} / \text { atom } \\ & =494 \mathrm{~kJ} / \mathrm{mol} \end{aligned}$

$\begin{aligned} & 5^{\text {th }} \text { excited state } \Rightarrow n_1=6 \\ & 1^{\text {st }} \text { excited state } \Rightarrow n_2=2 \\ & \Delta n=n_1-n_2=6-2=4 \end{aligned}$

Maximum number of spectral lines

$=\frac{\Delta \mathrm{n}(\Delta \mathrm{n}+1)}{2}=\frac{4(4+1)}{2}=10$

To determine the number of electrons with a spin quantum number of $s=+\frac{1}{2}$ in all completely filled subshells with principal quantum number $n=4$, we must first identify the subshells in the n=4 shell and then calculate the electrons with the specified spin.

The n=4 shell has the following subshells and their capacity for electrons:

• $4s$ subshell can hold 2 electrons
• $4p$ subshell can hold 6 electrons
• $4d$ subshell can hold 10 electrons
• $4f$ subshell can hold 14 electrons

In quantum mechanics, each orbital within these subshells can hold 2 electrons, each with opposite spins ($+\frac{1}{2}$ and $-\frac{1}{2}$). Thus, in a completely filled subshell, half of the electrons will have a spin quantum number of $+\frac{1}{2}$:

• In $4s$, 1 out of 2 electrons will have $s=+\frac{1}{2}$
• In $4p$, 3 out of 6 electrons will have $s=+\frac{1}{2}$
• In $4d$, 5 out of 10 electrons will have $s=+\frac{1}{2}$
• In $4f$, 7 out of 14 electrons will have $s=+\frac{1}{2}$

Adding these together: $1 + 3 + 5 + 7 = 16$ electrons with $s=+\frac{1}{2}$ in all completely filled subshells with $n=4$.

The photoelectric effect occurs when a photon with energy greater than the work function of a metal strikes the metal surface, causing electrons to be emitted. The minimum energy required for this process is given by the equation:

$\begin{equation} E = h\nu = \frac{hc}{\lambda} \end{equation}$

where $E$ is the energy of the photon, $h$ is Planck's constant, $\nu$ is the frequency of the photon, $c$ is the speed of light, $\lambda$ is the wavelength of the photon.

If the energy of the photon is greater than the work function ($W_0$) of the metal, then electrons will be emitted. The energy of the photon is related to its wavelength by the above equation.

For the given wavelength of 400 nm, the energy of the photon is:

$\begin{equation} E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \text{ J s} \times 3.0 \times 10^8 \text{ m/s}}{400 \times 10^{-9} \text{ m}} = 4.97 \times 10^{-19} \text{ J} \end{equation}$

For the photoelectric effect to occur, this energy must be greater than the work function of the metal. The metals that will show the photoelectric effect can be determined by comparing the energy of the photon to the work function of each metal.

From the table, we see that three metals (Li, Na, and K) have work functions less than the energy of the photon with a wavelength of 400 nm. Therefore, these three metals will show the photoelectric effect when light of this wavelength falls on them.

Thus, the number of metals which will show photoelectric effect when light of wavelength 400 nm falls on them is 3.

Let's look at each statement individually:

A. For 1s orbital, the probability density is maximum at the nucleus.

This statement is correct. For a 1s orbital, the electron is most likely to be found closest to the nucleus.

B. For 2s orbital, the probability density first increases to maximum and then decreases sharply to zero.

This statement is incorrect. For a 2s orbital, the probability density is maximum at the nucleus, then decreases to zero (this is the radial node), and then rises again to a secondary peak before gradually tapering off.

C. Boundary surface diagrams of the orbitals encloses a region of 100% probability of finding the electron.

This statement is incorrect. The boundary surface usually encloses a region where there is approximately a 90% chance of finding the electron, not 100%.

D. p and d-orbitals have 1 and 2 angular nodes respectively.

This statement is correct. The number of angular nodes corresponds to the orbital angular momentum quantum number (l). For a p-orbital, l=1, thus there is one angular node. For a d-orbital, l=2, thus there are two angular nodes.

E. Probability density of p-orbital is zero at the nucleus.

This statement is correct. The probability density of a p-orbital is indeed zero at the nucleus because p-orbitals have one angular node at the nucleus.

So, the total number of correct statements is 3 (A, D, E).

The energy levels of a hydrogen-like atom (such as Li²⁺ in this case) can be calculated using the formula:

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

where:

• $E$ is the energy of the electron,
• $R_H$ is the Rydberg constant (in joules),
• $Z$ is the atomic number, and
• $n$ is the principal quantum number.

The change in energy (∆E) between two energy levels is given by the difference in the energies of the two levels. In this case, the electron is excited from the nth level to the (n+1)th level, and the energy of the radiation is given as $1.47 \times 10^{-17}$ J. Therefore:

$\Delta E = E_{n+1} - E_n = 1.47 \times 10^{-17} \mathrm{~J}$

Substituting the formula for $E_{n}$ and $E_{n+1}$:

$\Delta E = -R_H \cdot \frac{Z^2}{(n+1)^2} - \left(-R_H \cdot \frac{Z^2}{n^2}\right)$

Simplifying this, we get:

$\Delta E = R_H \cdot Z^2 \left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right)$

We know that $Z=3$ for $\mathrm{Li}^{2+}$, $R_H = 2.18 \times 10^{-18} \mathrm{~J}$ and $\Delta E = 1.47 \times 10^{-17} \mathrm{~J}$.

We substitute the given values into the formula

$1.47 \times 10^{-17} \mathrm{~J} = 2.18 \times 10^{-18} \mathrm{~J} \times 3^2 \left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right)$

Simplify to

$\frac{1.47}{1.96} = \frac{1}{n^2} - \frac{1}{(n+1)^2}$

Solving this quadratic equation for n, we get two possible solutions. However, we disregard the negative solution since n (the principal quantum number) can't be negative. So, the only valid solution is n=1.

Thus, the value of n is 1.

All the statements (A, B, C, D) provided about black bodies are correct.

(A) A black body is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. It also emits energy in the form of electromagnetic radiation.

(B) The frequency distribution of the emitted radiation by a black body indeed depends on its temperature. This phenomenon is described by Planck's law.

(C) The intensity vs frequency curve for a black body, also known as the Planck's curve, does indeed pass through a maximum value for a given temperature.

(D) The frequency at which the maximum intensity occurs increases with the temperature of the black body. This is stated in Wien's displacement law.

So, the number of incorrect statements about the black body from the above is 0.

Radial nodes in an atomic orbital are areas where the probability of finding an electron is zero. The number of radial nodes in an orbital is given by the formula:

$ \text{number of radial nodes} = n - l - 1 $

where $n$ is the principal quantum number and $l$ is the azimuthal quantum number. The azimuthal quantum number ($l$) can have values from 0 to $n-1$, and it determines the shape of the orbital (s, p, d, f, etc.). For an s orbital, $l=0$; for a p orbital, $l=1$; for a d orbital, $l=2$; and so on.

Let's calculate the number of radial nodes for each given orbital:

1. 7s: $n=7$, $l=0$, so the number of radial nodes is $7 - 0 - 1 = 6$, not 5.
2. 7p: $n=7$, $l=1$, so the number of radial nodes is $7 - 1 - 1 = 5$.
3. 6s: $n=6$, $l=0$, so the number of radial nodes is $6 - 0 - 1 = 5$.
4. 8p: $n=8$, $l=1$, so the number of radial nodes is $8 - 1 - 1 = 6$, not 5.
5. 8d: $n=8$, $l=2$, so the number of radial nodes is $8 - 2 - 1 = 5$.

Therefore, the orbitals with 5 radial nodes are 7p, 6s, and 8d, so there are 3 such orbitals.

(A) This statement is correct. Line emission spectra (or atomic emission spectra) are indeed used to study the electronic structure of atoms. These spectra are produced when photons are emitted from atoms as excited electrons return to a lower energy level. Each line corresponds to a particular quantum leap between energy levels.

(B) This statement is incorrect. The emission spectra of atoms in the gas phase do not show a continuous spread of wavelength from red to violet. Instead, they show distinct lines corresponding to specific electron transitions, hence they are known as line emission spectra.

(C) This statement is correct. An absorption spectrum is like the photographic negative of an emission spectrum. In an absorption spectrum, light passes through a cold, dilute gas and atoms in the gas absorb at characteristic frequencies; since the re-emitted light is unlikely to be emitted in the same direction as the absorbed photon, this gives rise to dark lines (absence of light) in the spectrum.

(D) This statement is correct. The element helium was indeed discovered in the sun through spectroscopy before it was found on Earth. The name helium comes from Helios, the Greek name for the sun.

So, the number of incorrect statements is 1.

$\mathrm{E=nhv}$

$=(6.022\times10^{23})(6.626\times10^{-34})\times(2\times10^{12})$

$=798.03$ J

$\approx$ $798$ J

Radius of 3^rd Bohr orbit of He$^{+3}$

$ = 0.6 \times {{{{(3)}^2}} \over 2}$

$ = 0.3 \times 9$

$ = 2.7\,\mathop A\limits^o $

$ = 270 \times {10^{ - 12}}$ ppm

Given,

Work function = 6.63 $\times$ 10^$-$19 J

$ = {{6.63 \times {{10}^{ - 19}}} \over {1.6 \times {{10}^{ - 19}}}}$

= 4.14 eV

We know,

$E = {{1240} \over {\lambda \,(nm)}}$

$ \Rightarrow 4.14 = {{1240} \over \lambda }$

$\lambda$ = 300 nm

For A,

Given n = 3 and l = 3

but we know maximum value of l = n $-$ 1.

$\therefore$ l can't be equal to n.

So, Set A of quantum numbers is not possible.

For B,

Given n = 3, l = 2, m = $-$ 2

Here, l = 2 which follow the rule l = n $-$ 1.

And we know possible value of m is $-$ l to + l.

here possible value of m = $-$2 to +2

$\therefore$ This Set B is valid set of quantum numbers.

For C,

Given n = 2, l = 1, m = +1

Here l = 1 which follows the rule l = n $-$ 1.

For l = 1 possible value of m = $-$1 to +1

Here m = +1. So value of m is valid.

$\therefore$ Set C is valid set of quantum numbers.

For D,

Given n = 2, l = 2, m = +2

l = 2 does not follow the rule l = n $-$ 1 rule.

$\therefore$ Set D is not valid set of quantum numbers.

We know from hisenberg uncertainty principle

$\Delta x\,.\,\Delta p = {h \over {4\pi }}$

$ \Rightarrow \Delta x\,.\,m\Delta v = {h \over {4\pi }}$

Given,

$\Delta$x = 10^$-$7 m

$\Delta$x = 2.4 $\times$ 10^$-$26 m/s

h = 6.626 $\times$ 10^$-$34 Js

$\therefore$ ${10^{ - 7}} \times m \times 2.4 \times {10^{ - 26}} = {{6.626 \times {{10}^{ - 34}}} \over {4\pi }}$

$ \Rightarrow 2.4m = {{6.626} \over {4\pi \times 10}}$

$\Rightarrow$ m = 0.022 kg

$\Rightarrow$ m = 22 gm

Bohr model is not valid for lithium atom (Li) as Bohr model is valid for only single electronic species, so it would be valid for Li⁺² but not Li atom.

So this question is BONUS.