Structure of Atom
A hypothetical electromagnetic wave is show below.
The frequency of the wave is $\mathrm{x} \times 10^{19} \mathrm{~Hz}$.
$\mathrm{x}=$ _________ (nearest integer)
Explanation:
$\begin{aligned} \lambda & =1.5 \times 4=6 \mathrm{~pm} \\ \nu & =\frac{C}{\lambda} \\ \nu & =\frac{3 \times 10^8}{6 \times 10^{-12}} \\ & =0.5 \times 10^{20} \\ \nu & =5 \times 10^{19} \\ x & =5 \end{aligned}$
For hydrogen atom, energy of an electron in first excited state is $-3.4 \mathrm{~eV}, \mathrm{K} . \mathrm{E}$. of the same electron of hydrogen atom is $x \mathrm{~eV}$. Value of $x$ is _________ $\times 10^{-1} \mathrm{~eV}$. (Nearest integer)
Explanation:
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.
Frequency of the de-Broglie wave of electron in Bohr's first orbit of hydrogen atom is _________ $\times 10^{13} \mathrm{~Hz}$ (nearest integer).
[Given : $\mathrm{R}_{\mathrm{H}}$ (Rydberg constant) $=2.18 \times 10^{-18} \mathrm{~J}, h$ (Plank's constant) $=6.6 \times 10^{-34} \mathrm{~J} . \mathrm{s}$.]
Explanation:
$\begin{aligned} & \lambda=\frac{h}{m v} \\ & \lambda \cdot v=\frac{h}{m} \\ & \frac{\lambda \cdot v^2}{v}=\frac{h}{m} \\ & \frac{v^2}{\text { Frequency }}=\frac{h}{m} \end{aligned}$
$\begin{aligned} & \text { Frequency }=\frac{\mathrm{mv}^2}{\mathrm{~h}}=\frac{2 \mathrm{R}_{\mathrm{H}}}{\mathrm{h}} \\ & =\frac{2 \times 2.18 \times 10^{-18}}{6.6 \times 10^{-34}} \\ & =660.6 \times 10^{13} \mathrm{~Hz} \end{aligned}$
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.
In an atom, total number of electrons having quantum numbers $\mathrm{n}=4,\left|\mathrm{~m}_l\right|=1$ and $\mathrm{m}_{\mathrm{s}}=-\frac{1}{2}$ is _________.
Explanation:
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 ml 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.
The value of Rydberg constant $(R_H)$ is $2.18 \times 10^{-18} \mathrm{~J}$. The velocity of electron having mass $9.1 \times 10^{-31} \mathrm{~kg}$ in Bohr's first orbit of hydrogen atom = ________ $\times 10^5 \mathrm{~ms}^{-1}$ (nearest integer).
Explanation:
$\begin{aligned} & \text { K.E. }=R_H \cdot \frac{z^2}{n^2}=\frac{1}{2} m v^2 \\ & v^2=\frac{2 \times 2.18 \times 10^{-18}}{9.1 \times 10^{-31}} \times \frac{1}{1}=0.479 \times 10^{13} \\ & v=21.88 \times 10^5 \mathrm{~m} / \mathrm{s} \end{aligned}$
The maximum number of orbitals which can be identified with $\mathrm{n}=4$ and $m_l=0$ is _________.
Explanation:
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's wavelength of an electron in the $4^{\text {th }}$ orbit is ________ $\pi \mathrm{a}_0$. ($\mathrm{a}_0=$ Bohr's radius)
Explanation:
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 $
The ionization energy of sodium in $\mathrm{~kJ} \mathrm{~mol}^{-1}$, if electromagnetic radiation of wavelength $242 \mathrm{~nm}$ is just sufficient to ionize sodium atom is _______.
Explanation:
$\begin{aligned} & \mathrm{E}=\frac{1240}{\lambda(\mathrm{nm})} \mathrm{eV} \\ & =\frac{1240}{242} \mathrm{eV} \\ & =5.12 \mathrm{eV} \\ & =5.12 \times 1.6 \times 10^{-19} \\ & =8.198 \times 10^{-19} \mathrm{~J} / \text { atom } \\ & =494 \mathrm{~kJ} / \mathrm{mol} \end{aligned}$
Number of spectral lines obtained in $\mathrm{He}^{+}$ spectra, when an electron makes transition from fifth excited state to first excited state will be
Explanation:
$\begin{aligned} & 5^{\text {th }} \text { excited state } \Rightarrow n_1=6 \\ & 1^{\text {st }} \text { excited state } \Rightarrow n_2=2 \\ & \Delta n=n_1-n_2=6-2=4 \end{aligned}$
Maximum number of spectral lines
$=\frac{\Delta \mathrm{n}(\Delta \mathrm{n}+1)}{2}=\frac{4(4+1)}{2}=10$
The number of electrons present in all the completely filled subshells having $\mathrm{n}=4$ and $\mathrm{s}=+\frac{1}{2}$ is _______.
(Where $\mathrm{n}=$ principal quantum number and $\mathrm{s}=$ spin quantum number)
Explanation:
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 $n=4$.
Among the following, the correct statement(s) for electrons in an atom is(are)
Statement I : According to Bohr's model of hydrogen atom, the angular momentum of an electron in a given stationary state is quantised.
Statement II : The concept of electron in Bohr's orbit, violates the Heisenberg uncertainty principle.
In the light of the above statements, choose the most appropriate answer from the options given below:
The energy of an electron in the first Bohr orbit of hydrogen atom is $-2.18 \times 10^{-18} \mathrm{~J}$. Its energy in the third Bohr orbit is ____________.
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R :
Assertion A : In the photoelectric effect, the electrons are ejected from the metal surface as soon as the beam of light of frequency greater than threshold frequency strikes the surface.
Reason R : When the photon of any energy strikes an electron in the atom, transfer of energy from the photon to the electron takes place.
In the light of the above statements, choose the most appropriate answer from the options given below :
Henry Moseley studied characteristic X-ray spectra of elements. The graph which represents his observation correctly is
Given $v=$ frequency of $\mathrm{X}$-ray emitted
Z = atomic number
If the radius of the first orbit of hydrogen atom is $\alpha_{0}$, then de Broglie's wavelength of electron in $3^{\text {rd }}$ orbit is :
Which one of the following sets of ions represents a collection of isoelectronic species?
(Given : Atomic Number : $\mathrm{F:9,Cl:17,Na=11,Mg=12,Al=13,K=19,Ca=20,Sc=21}$)
A. $\mathrm{n}=3, \mathrm{l}=0, \mathrm{~m}=0$
B. $\mathrm{n}=4, \mathrm{l}=0, \mathrm{~m}=0$
C. $\mathrm{n}=3, \mathrm{l}=1, \mathrm{~m}=0$
D. $\mathrm{n}=3, \mathrm{l}=2, \mathrm{~m}=1$
The correct option for the order is :
Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from $\mathrm{n=4}$ to $\mathrm{n}=2$ of $\mathrm{He}^{+}$ spectrum
$ \Psi_{2 \mathrm{~s}}=\frac{1}{2 \sqrt{2 \pi}}\left(\frac{1}{a_0}\right)^{1 / 2}\left(2-\frac{r}{a_0}\right) e^{-r / 2 a_0} $
At $r=r_0$, radial node is formed. Thus, $r_0$ in terms of $a_0$
The shortest wavelength of hydrogen atom in Lyman series is $\lambda$. The longest wavelength is Balmer series of He$^+$ is
The radius of the $\mathrm{2^{nd}}$ orbit of $\mathrm{Li^{2+}}$ is $x$. The expected radius of the $\mathrm{3^{rd}}$ orbit of $\mathrm{Be^{3+}}$ is
The number of s-electrons present in an ion with 55 protons in its unipositive state is
The magnetic moment of a transition metal compound has been calculated to be 3.87 B.M. The metal ion is
It is observed that characteristic X-ray spectra of elements show regularity. When frequency to the power "n" i.e. ${v^n}$ of X-rays emitted is plotted against atomic number "Z", following graph is obtained.
The value of 'n' is :
$\mathrm{O}^{2-}, \mathrm{F}^{-}, \mathrm{Al}, \mathrm{Mg}^{2+}, \mathrm{Na}^{+}, \mathrm{O}^{+}, \mathrm{Mg}, \mathrm{Al}^{3+}, \mathrm{F}$
Explanation:
The electron configuration of each species is:
- $\mathrm{O}^{2-}$: $1s^2\ 2s^2\ 2p^6$, 10 electrons
- $\mathrm{F}^{-}$: $1s^2\ 2s^2\ 2p^6$, 10 electrons
- $\mathrm{Al}$: $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^1$, 13 electrons
- $\mathrm{Mg}^{2+}$: $1s^2\ 2s^2\ 2p^6$, 10 electrons
- $\mathrm{Na}^{+}$: $1s^2\ 2s^2\ 2p^6$, 10 electrons
- $\mathrm{O}^{+}$: $1s^2\ 2s^2\ 2p^3$, 7 electrons
- $\mathrm{Mg}$: $1s^2\ 2s^2\ 2p^6\ 3s^2$, 12 electrons
- $\mathrm{Al}^{3+}$: $1s^2\ 2s^2\ 2p^6$, 10 electrons
- $\mathrm{F}$: $1s^2\ 2s^2\ 2p^5$, 9 electrons
The orbital angular momentum of an electron in $3 \mathrm{~s}$ orbital is $\frac{x h}{2 \pi}$. The value of $x$ is ____________ (nearest integer)
Explanation:
$L = \sqrt{l(l+1)} \frac{h}{2\pi}$
Where $l$ is the azimuthal quantum number (orbital angular momentum quantum number) and $h$ is the Planck's constant.
For a 3s orbital, the principal quantum number (n) is 3 and the azimuthal quantum number (l) is 0, as s orbitals have l=0. Now, let's calculate the orbital angular momentum:
$L = \sqrt{0(0+1)} \frac{h}{2\pi} = 0$
Thus, the orbital angular momentum of an electron in a 3s orbital is 0. So, the value of $x$ is 0.
Values of work function (W$_0$) for a few metals are given below
| Metal | Li | Na | K | Mg | Cu | Ag |
|---|---|---|---|---|---|---|
| W$_0$/eV | 2.42 | 2.3 | 2.25 | 3.7 | 4.8 | 4.3 |
The number of metals which will show photoelectric effect when light of wavelength $400 \mathrm{~nm}$ falls on it is ___________
Given: $\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}$
$c=3 \times 10^{8} \mathrm{~ms}^{-1}$
$e=1.6 \times 10^{-19} \mathrm{C}$
Explanation:
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.
| Metal | $W_0$ (eV) | $W_0$ (J) | $E > W_0$? |
|---|---|---|---|
| Li | 2.42 | 3.88 $\times$ 10$^{-19}$ J | Yes |
| Na | 2.3 | 3.68 $\times$ 10$^{-19}$ J | Yes |
| K | 2.25 | 3.60 $\times$ 10$^{-19}$ J | Yes |
| Mg | 3.7 | 5.92 $\times$ 10$^{-19}$ J | No |
| Cu | 4.8 | 7.68 $\times$ 10$^{-19}$ J | No |
| Ag | 4.3 | 6.88 $\times$ 10$^{-19}$ J | No |
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.
The number of correct statements from the following is ___________.
A. For $1 \mathrm{s}$ orbital, the probability density is maximum at the nucleus
B. For $2 \mathrm{s}$ orbital, the probability density first increases to maximum and then decreases sharply to zero.
C. Boundary surface diagrams of the orbitals encloses a region of $100 \%$ probability of finding the electron.
D. p and d-orbitals have 1 and 2 angular nodes respectively.
E. probability density of p-orbital is zero at the nucleus
Explanation:
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 electron in the $\mathrm{n}^{\text {th }}$ orbit of $\mathrm{Li}^{2+}$ is excited to $(\mathrm{n}+1)$ orbit using the radiation of energy $1.47 \times 10^{-17} \mathrm{~J}$ (as shown in the diagram). The value of $\mathrm{n}$ is ___________
Given: $\mathrm{R}_{\mathrm{H}}=2.18 \times 10^{-18} \mathrm{~J}$
Explanation:
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.
The number of incorrect statement/s about the black body from the following is __________
(A) Emit or absorb energy in the form of electromagnetic radiation.
(B) Frequency distribution of the emitted radiation depends on temperature.
(C) At a given temperature, intensity vs frequency curve passes through a maximum value.
(D) The maximum of the intensity vs frequency curve is at a higher frequency at higher temperature compared to that at lower temperature.
Explanation:
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.
The number of atomic orbitals from the following having 5 radial nodes is ___________.
$7 \mathrm{s}, 7 \mathrm{p}, 6 \mathrm{s}, 8 \mathrm{p}, 8 \mathrm{d}$
Explanation:
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:
- 7s: $n=7$, $l=0$, so the number of radial nodes is $7 - 0 - 1 = 6$, not 5.
- 7p: $n=7$, $l=1$, so the number of radial nodes is $7 - 1 - 1 = 5$.
- 6s: $n=6$, $l=0$, so the number of radial nodes is $6 - 0 - 1 = 5$.
- 8p: $n=8$, $l=1$, so the number of radial nodes is $8 - 1 - 1 = 6$, not 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.
The number of following statement/s which is/are incorrect is ___________
(A) Line emission spectra are used to study the electronic structure
(B) The emission spectra of atoms in the gas phase show a continuous spread of wavelength from red to violet
(C) An absorption spectrum is like the photographic negative of an emission spectrum
(D) The element helium was discovered in the sun by spectroscopic method
Explanation:
(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.
The wavelength of an electron of kinetic energy $4.50\times10^{-29}$ J is _________ $\times 10^{-5}$ m. (Nearest integer)
Given : mass of electron is $9\times10^{-31}$ kg, h $=6.6\times10^{-34}$ J s
Explanation:
Therefore Ans $=7$
Electrons in a cathode ray tube have been emitted with a velocity of 1000 m s$^{-1}$. The number of following statements which is/are $\underline {\mathrm{true}} $ about the emitted radiation is ____________.
Given : $\mathrm{h=6\times10^{-34}~J~s,m_e=9\times10^{-31}~kg}$.
(A) The de-Broglie wavelength of the electron emitted is 666.67 nm.
(B) The characteristic of electrons emitted depend upon the material of the electrodes of the cathode ray tube.
(C) The cathode rays start from cathode and move towards anode.
(D) The nature of the emitted electrons depends on the nature of the gas present in cathode ray tube.
Explanation:
(A) is True: The de-Broglie wavelength of an electron can be calculated as lambda = h / p, where h is Planck's constant and p is the momentum of the electron. The momentum of an electron with a velocity of 1000 m/s is mv, where m is the mass of an electron. Plugging in the values, we get
$\begin{aligned} & \mathrm{V}_{\mathrm{e}}=1000 \mathrm{~m} / \mathrm{s} ; \mathrm{h}=6 \times 10^{-34} \mathrm{Js} \\\\ & \mathrm{m}_{\mathrm{e}}=9 \times 10^{-31} \mathrm{~kg} \\\\ & \lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6 \times 10^{-34}}{9 \times 10^{-31} \times 1000}=666.67 \times 10^{-9} \mathrm{~m} \\\\ & =666.67 \mathrm{~nm}\end{aligned}$,
(B) is False : The characteristic of electrons emitted is independent of the material of the electrodes of the cathode ray tube.
(C) is True: In a cathode ray tube, electrons are emitted from the cathode and move towards the anode under the influence of an electric field.
(D) is False: The nature of the emitted electrons does not depend on the nature of the gas present in the cathode ray tube, but rather on the material of the electrodes and the potential difference applied across them.
The energy of one mole of photons of radiation of frequency $2 \times 10^{12} \mathrm{~Hz}$ in $\mathrm{J} ~\mathrm{mol}^{-1}$ is ___________. (Nearest integer)
[Given : $\mathrm{h}=6.626 \times 10^{-34} ~\mathrm{Js}$
$\mathrm{N}_{\mathrm{A}}=6.022 \times 10^{23} \mathrm{~mol}^{-1}$]
Explanation:
$\mathrm{E=nhv}$
$=(6.022\times10^{23})(6.626\times10^{-34})\times(2\times10^{12})$
$=798.03$ J
$\approx$ $798$ J
Assume that the radius of the first Bohr orbit of hydrogen atom is 0.6 $\mathrm{\mathop A\limits^o }$. The radius of the third Bohr orbit of He$^+$ is __________ picometer. (Nearest Integer)
Explanation:
Radius of 3rd 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
The number of given orbitals which have electron density along the axis is _________
$\mathrm{p_x,p_y,p_z,d_{xy},d_{yz},d_{xz},d_{z^2},d_{x^2-y^2}}$
Explanation:
Therefore, the total number of given orbitals which have electron density along the axis is 5.
If wavelength of the first line of the Paschen series of hydrogen atom is 720 nm, then the wavelength of the second line of this series is _________ nm. (Nearest integer)
Explanation:
$ \begin{aligned} & \Rightarrow R=\frac{9 \times 16}{720 \times 7} \\\\ & \frac{1}{\lambda^{\prime}}=\frac{9 \times 16}{720 \times 7} \times\left(\frac{1}{9}-\frac{1}{25}\right) \\\\ & \lambda^{\prime}=492.18 \mathrm{~nm} \\\\ & \lambda^{\prime}=492 \mathrm{~nm} \text { (nearest integer) } \end{aligned} $
[Use :
Bohr radius, $\mathrm{a}=52.9 \mathrm{pm}$
Rydberg constant, $R_{\mathrm{H}}=2.2 \times 10^{-18} \mathrm{~J}$
Planck's constant, $\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}$
Speed of light, $\mathrm{c}=3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}$ ]
Explanation:
$r = \frac{52.9 \times n^2}{Z}$
where $r$ is in pm, $n$ is the principal quantum number, and $Z$ is the atomic number.
2. For a $\mathrm{He}^{+}$ ion, $Z=2$.
3. We are given the initial radius $r_2 = 105.8$ pm and the final radius $r_1 = 26.45$ pm.
4. We can substitute these radii into the equation for $r$ to find the corresponding quantum numbers.
For $r_2 = 105.8$ pm, we get :
$105.8 = \frac{52.9 \times n_2^2}{2}$
which gives $n_2 = 2$.
Similarly, for $r_1 = 26.45$ pm, we get :
$26.45 = \frac{52.9 \times n_1^2}{2}$
which gives $n_1 = 1$.
5. Therefore, the transition is from $n_2 = 2$ to $n_1 = 1$.
6. The energy difference during this transition is equal to the energy of the emitted photon, which is given by :
$E = \frac{hc}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
where $h$ is the Planck's constant, $c$ is the speed of light, $R_H$ is the Rydberg constant, and $\lambda$ is the wavelength of the photon.
7. Substituting all known values into this equation gives :
$\frac{6.6 \times 10^{-34} \, \mathrm{J} \, \mathrm{s} \times 3 \times 10^8 \, \mathrm{m/s}}{\lambda} = 2.2 \times 10^{-18} \, \mathrm{J} \times 2^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right)$
Solving this equation yields $\lambda = 30 \times 10^{-9}$ m, or 30 nm.
8. So, the wavelength of the emitted photon during the transition is 30 nm.
Given below are the quantum numbers for 4 electrons.
A. $\mathrm{n}=3,l=2, \mathrm{~m}_{1}=1, \mathrm{~m}_{\mathrm{s}}=+1 / 2$
B. $\mathrm{n}=4,l=1, \mathrm{~m}_{1}=0, \mathrm{~m}_{\mathrm{s}}=+1 / 2$
C. $\mathrm{n}=4,l=2, \mathrm{~m}_{1}=-2, \mathrm{~m}_{\mathrm{s}}=-1 / 2$
D. $\mathrm{n}=3,l=1, \mathrm{~m}_{1}=-1, \mathrm{~m}_{\mathrm{s}}=+1 / 2$
The correct order of increasing energy is :
Given below are two statements: One is labelled as Assertion $\mathbf{A}$ and the other is labelled as Reason $\mathbf{R}$
Assertion $\mathbf{A}$ : Zero orbital overlap is an out of phase overlap.
Reason $\mathbf{R}$ : It results due to different orientation / direction of approach of orbitals.
In the light of the above statements, choose the correct answer from the options given below
Identify the incorrect statement from the following.
The correct decreasing order of energy for the orbitals having, following set of quantum numbers :
(A) n = 3, l = 0, m = 0
(B) n = 4, l = 0, m = 0
(C) n = 3, l = 1, m = 0
(D) n = 3, l = 2, m = 1
is :
Outermost electronic configurations of four elements A, B, C, D are given below :
(A) $3 s^{2}$
(B) $3 s^{2} 3 p^{1}$
(C) $3 s^{2} 3 p^{3}$
(D) $3 s^{2} 3 p^{4}$
The correct order of first ionization enthalpy for them is :
Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: Energy of $2 \mathrm{s}$ orbital of hydrogen atom is greater than that of $2 \mathrm{s}$ orbital of lithium.
Reason R: Energies of the orbitals in the same subshell decrease with increase in the atomic number.
In the light of the above statements, choose the correct answer from the options given below.
Which of the following sets of quantum numbers is not allowed?
The number of radial nodes and total number of nodes in 4p orbital respectively are :