M.C.Q (1 Marks) - Physics STD 11 Science Questions - Vidyadip

Questions

M.C.Q (1 Marks)

Take a timed test

18 questions · auto-graded multiple-choice test.

MCQ 11 Mark

The minimum orbital angular momentum of the electron in a hydrogen atom is:

A
$\text{h}$
B
$\frac{\text{h}}{2}$
✓
$\frac{\text{h}}{2\pi}$
D
$\frac{\text{h}}{\lambda}$

Answer

Correct option: C.

$\frac{\text{h}}{2\pi}$

According to Bohr's atomic theory, the orbital angular momentum of an electron is an integral multiplt of $\frac{\text{h}}{2\pi}$
$\therefore\ \text{L}_\text{n}=\frac{\text{nh}}{2\pi}$
Here,
n = Principal quantum number
The minimum of n is 1
Thus, the minimum value of the orbit angular momentum of the electron in a hydrogen is given by $\text{L}=\frac{\text{h}}{2\pi}$

View full question & answer→

MCQ 21 Mark

In which of the following systems will the wavelength corresponding to $n = 2$ to $n = 1$ be minimum?

A
Hydrogen atom.
B
Deuterium atom.
C
Singly ionized helium.
✓
Doubly ionized lithium.

Answer

Correct option: D.

Doubly ionized lithium.

The wavelength corresponding the transition from $n_2$ to $n_1$ is given by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\Big)$
Here,
$R =$ Rydberg constant.
$Z =$ Atomic number of the ion.
From the given formula, it can be observed that the wavelength is inversely proportional to the square of the atomic number.
Therefore, the wavelength corresponding to $n = 2$ to $n = 1$ will be minimum in doubly ionized lithium ion because for lithium, $Z = 3.$

View full question & answer→

MCQ 31 Mark

In a laser tube, all the photons:

A
Have same wavelength.
B
Have same energy.
C
Move in same direction.
✓
Move with same speed.

Answer

Correct option: D.

Move with same speed.

All the photons emitted in the laser move with the speed equal to the speed of light $(c = 3 \times 10^8m/s).$ Ideally, the light wave through the laser must be coherent, but in practical laser tubes, there is some deviation from the ideal result. Thus, the photons emitted by the laser have little variations in their wavelengths and energies as well as the directions, but the velocity of all the photons remains same.

View full question & answer→

MCQ 41 Mark

Let $A_n$ be the area enclosed by the $n^{th}$ orbit in a hydrogen atom. The graph of ln $\Big(\frac{\text{A}_\text{n}}{\text{A}_1}\Big)$ against in $(n):$

A
Will pass through the origin.
B
Will be a straight line with slope $4$.
✓
$A$ and $B$
D
Will be a circle.

Answer

Correct option: C.

$A$ and $B$

$\text{r}_\text{n}=\text{n}^2\text{a}_0$
Area of the nth orbit is given by,
$\text{A}_\text{n}=\pi\text{r}^2_\text{n}=\pi\text{n}^4\text{a}^2_0$
$\text{A}_1=\pi\text{a}^{2}_0$
$\text{ln}\Big(\frac{\text{A}_\text{n}}{\text{A}_1}\Big)=\text{ln}\Big(\frac{\pi\text{n}^4\text{a}_0^2}{\pi\text{a}^2_0}\Big)$
$\text{ln}\Big(\frac{\text{A}_\text{n}}{\text{A}_1}\Big)=4\text{ln n}\ ...(\text{i})$
From the above expression, the graph of in $\Big(\frac{\text{A}_\text{n}}{\text{A}_1}\Big)$ against ln $(n)$ will be a straight line passing through the origin and having slope $4$.

View full question & answer→

MCQ 51 Mark

Which of the following products in a hydrogen atom are independent of the principal quantum number $n?$ The symbols have their usual meanings.

A
$vn.$
✓
$Er.$
C
$A$ and $B$
D
$vr.$

Answer

Correct option: B.

$Er.$

Relations for energy, radius of the orbit and its velocity are given by,
$\text{E}=-\frac{\text{mZ}^2e^4}{8\in_0^2\text{h}^2\text{n}^2}$
$\text{r}=\frac{\in_0\text{h}^2\text{n}^2}{\pi\text{mZe}^2}$
$\text{v}=\frac{\text{Ze}^2}{2\in_0\text{hn}}$
Where,
$Z:$ The atomic number of hydrogen like atom.
$e:$ Electric charge.
$h:$ Plank constant.
$m:$ Mass of electron.
$n:$ Principal quantam number of the electron.
$\in_0:$ Permittivity of vacuum
From these relations, we can see that the products independent of $n$ are $vn, Er.$

View full question & answer→

MCQ 61 Mark

Three photons coming from excited atomic-hydrogen sample are picked up. Their energies are 12.1eV, 10.2 eV and 1.9eV. These photons must come from:

A
A single atom.
B
Two atoms.
C
Three atoms.
✓
Either two atoms or three atoms.

Answer

Correct option: D.

Either two atoms or three atoms.

The energies of the photons emitted can be expressed as follows,

Energy of photon	Transition
12.1eV	n = 3 to n = 1
10.2eV	n = 2 to n = 1
1.9eV	n = 3 to n = 2

A hydrogen atom consists of only one electron. An electron can have transitions, like from n = 3 to n = 2 or from n = 2 to n = 1, at a time.

So, it can be concluded that the photons are emitted either from three atoms (when all the three transitions of electrons are in different atoms) or from two atoms (when an atom has n = 3 to n = 2 and then n = 2 to n = 1 electronic transition and the other has n = 3 to n = 1 electronic transition).

View full question & answer→

MCQ 71 Mark

In which of the following systems will the radius of the first orbit $(n = 1)$ be minimum?

A
Hydrogen atom.
B
Deuterium atom.
C
Singly ionized helium.
✓
Doubly ionized lithium.

Answer

Correct option: D.

Doubly ionized lithium.

For a hydrogen$-$like ion with $Z$ protons in the nucleus, the radius of the $\ce{n^{th}}$ state is given by,
$\text{r}_\text{n}=\frac{\text{n}^2\text{a}_0}{\text{Z}}$
Here,
$a_0 = 0.53$ pm
For lithium, $Z = 3$ Therefore, the radius of the first orbit for doubly ionised lithium will be minimum.

View full question & answer→

MCQ 81 Mark

Ionization energy of a hydrogen$-$like ion $A$ is greater than that of another hydrogen$-$like ion $B$. Let $r, u, E$ and $L$ represent the radius of the orbit, speed of the electron, energy of the atom and orbital angular momentum of the electron respectively. In ground state:

A
$r_A>r_B$
✓
$u_A>u_B$
C
$E_A>E_B$
D
$L_A>L_B$

Answer

Correct option: B.

$u_A>u_B$

The ionisation energy of a hydrogen like ion of atomic number $Z$ is given by,
$\text{V}=(13.6\text{eV})\times\text{Z}^2$
Thus, the atomic number of ion $A$ is greater than that of $B(Z_A > Z_B)$.
The radius of the orbit is inversely proportional to the atomic number of the ion.
$\therefore\ \text{r}_\text{A}>\text{r}_\text{B}$
The speed of electron is directly proportional to the atomic number.
Therefore, the speed of the electron in the orbit of $A$ will be more than that in $B.$
Thus, $u_A > u_B$ is correct.
The total energy of the atom is given by,
$\text{E}=-\frac{\text{mZ}^2\text{e}^2}{8\in_0\text{h}^2\text{n}^2}$
As the energy is directly proportional to $Z^2$, the energy of A will be less than that of $B,$
i.e. $E_A < E_B$.
The orbital angular momentum of the electron is independent of the atomic number.
Therefore, the relation $L_A > L_B$ is invalid.

View full question & answer→

MCQ 91 Mark

In which of the following transitions will the wavelength be minimum?

A
n = 5 to n = 4
B
n = 4 to n = 3
C
n = 3 to n = 2
✓
n = 2 to n = 1

Answer

Correct option: D.

n = 2 to n = 1

For the transition in the hydrogen-like atom, the wavelength of the emitted radiation is calculated by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{\text{n}_1}-\frac{1}{\text{n}_2}\Big)$
Here, R is the Rydberg constant.
For the transition from n = 5 to n = 4, the wavelength is given by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{4^2}-\frac{1}{5^2}\Big)$
$\lambda=\frac{400}{9\text{RZ}^2}$
For the transition from n = 4 to n = 3, the wavelength is given by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{3^2}-\frac{1}{4^2}\Big)$
$\lambda=\frac{144}{7\text{RZ}^2}$
For the transition from n = 3 to n = 2, the wavelength is given by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{2^2}-\frac{1}{3^2}\Big)$
$\lambda=\frac{36}{5\text{RZ}^2}$
For the transition form n = 2 to n = 1, the wavelength is given by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{1^2}-\frac{1}{2^2}\Big)$
$\lambda=\frac{2}{\text{RZ}^2}$
From the above calculations, it can be observed that the wavelength of the radiation emitted for the transition from n = 2 to n = 1 will be minimum.

View full question & answer→

MCQ 101 Mark

The radius of the shortest orbit in a one$-$electron system is $18\ pm.$ It may be

A
Hydrogen.
B
Deuterium.
C
$He^+$
✓
$Li^{++}$

Answer

Correct option: D.

$Li^{++}$

The radius of the $n^{\text {th }}$ orbit in one electron system is given by,
$\mathrm{r}_{\mathrm{n}}=\frac{\mathrm{n}^2 \mathrm{a}_0}{\mathrm{Z}}$
Here, $a_0=53 \mathrm{pm}$
For the shortest orbit, $\mathrm{n}=1$
For hydrogen, $\mathrm{Z}=1$
$\therefore$ Radius of the first state of hydrogen atom $=53 \mathrm{pm}$
For deuterium, $\mathrm{Z}=1$
$\therefore$ Radius of the first state of deuterium atom $=53 \mathrm{pm}$
For $\mathrm{He}^{+}, \mathrm{Z}=2$
$\therefore$ Radius of $\mathrm{He}^{+}$atom $=\frac{53}{2} \mathrm{pm}=26.5 \mathrm{pm}$
For $\mathrm{Li}^{++}, \mathrm{Z}=3$
$\therefore$ Radius of $\mathrm{Li}^{++}$atom $=\frac{53}{3} \mathrm{pm}=17.66 \approx 18 \mathrm{pm}$
The given one$-$electron system having radius of the shortest orbit to be $18\ pm$ may be $\mathrm{Li}^{++}$

View full question & answer→

MCQ 111 Mark

An electron with kinetic energy $5eV$ is incident on a hydrogen atom in its ground state. The collision:

✓
Must be elastic.
B
May be partially elastic.
C
Must be completely inelastic.
D
May be completely inelastic.

Answer

Correct option: A.

Must be elastic.

The minimum energy required to excite a hydrogen atom from its ground state to $1^{st}$ excited state is approximately $10eV.$ As the incident electron energy is not sufficient for excitation of the hydrogen atom so electron will not get absorbed in the hydrogen atom so it can not be an inelastic collision. Also this collision can not be partially elastic because in an partially elastic collision, there is a net loss on kinetic energy. If the energy is lost then corresponding amount of heat should have been produced but it is not so which implies that the collision is completely elastic.

View full question & answer→

MCQ 121 Mark

The energy of an atom $($or ion$)$ in its ground state is $-54.4eV$. It may be:

A
Hydrogen.
B
Deuterium.
✓
$He^+$
D
$Li^{++}$

Answer

Correct option: C.

$He^+$

The total energy of a hydrogen$-$like ion, having $Z$ protons in its nucleus, is given by,
$\text{E}=-\frac{13.6\text{Z}^2}{\text{n}^2}\text{eV}$
Here, $n =$ Principal quantum number.
For ground state, $n = 1$
$\therefore$ Total energy, $E = -13.6Z^2eV$
For hydrogen, $Z = 1$
$\therefore$ Total energy, $E = -13.6eV$
For deuterium, $Z = 1$
$\therefore$ Total energy, $e = -13.6eV$
For $He^+, Z = 2$
$\therefore$ Total energy $E = -13.6 \times 2^2 = -54.4eV$
For $Li^{++}$,
$Z = 3$
$\therefore$ Total energy, $E = -13.6 \times 3^2 = -122.4eV$
Hence, the ion having an energy of $-54.4eV$ in its ground state may be $He^+$

View full question & answer→

MCQ 131 Mark

As one considers orbits with higher values of $n$ in a hydrogen atom, the electric potential energy of the atom:

A
Decreases.
✓
Increases.
C
Remains the same.
D
Does not increase.

Answer

Correct option: B.

Increases.

The electric potential energy of hydrogen atom with electron at the $n^{th}$ state is given by,
$\text{V}=-\frac{2\times13.6}{\text{n}^2}$
As the value of $n$ increases, the potential energy of the hydrogen atom also increases,
i.e. the atom becomes less bound as $n$ increases.

View full question & answer→

MCQ 141 Mark

When a photon stimulates the emission of another photon, the two photons have:

A
Same energy.
B
Same direction.
C
Same phase.
✓
All of the above

Answer

Correct option: D.

All of the above

When a photon stimulates the emission of another photon, the two photons have same energy, direction, phase, and wavelength or we can say that the two photons are coherent. When an atom is present in its excited state then if a photon of energy equal to the energy gap between the excited state and any lower stable state is incident on this atom then the atom transits from upper state to the lower stable state by emitting a photon of energy equal to the energy gap between the two states. It is called stimulated emission. The emitted photon and incident photon have same energy and hence same wavelength. Also these two photons will be in phase and in the same direction. This process of producing monochromatic and unidirectional light is called lasing action.

View full question & answer→

MCQ 151 Mark

Suppose, the electron in a hydrogen atom makes transition from $n = 3$ to $n = 2$ in $10^{-8}s$. The order of the torque acting on the electron in this period, using the relation between torque and angular momentum as discussed in the chapter on rotational mechanics is:

A
$10^{-34} \mathrm{Nm}$
B
$10^{-24} \mathrm{Nm}$
✓
$10^{-42} \mathrm{Nm}$
D
$10^{-8} \mathrm{Nm}$

Answer

Correct option: C.

$10^{-42} \mathrm{Nm}$

The angular momentum of the electron for the $n^{th}$ state is given by,
$\text{L}_\text{n}=\frac{\text{nh}}{2\pi}$
Angular momentum of the electron for $\text{n}=3,\ \text{L}_\text{i}=\frac{3\text{h}}{2\pi}$
Angular momentum of the electron for $\text{n}=2,\ \text{L}_\text{f}=\frac{2\text{h}}{2\pi}$
The torque is the time rate of change of the angular momentum.
Torque, $\tau=\frac{\text{L}_\text{f}-\text{L}_\text{i}}{\text{t}}$
$=\frac{\big(\frac{2\text{h}}{2\pi}\big)-\big(\frac{3\text{h}}{2\pi}\big)}{10^{-8}}$
$=\frac{-\big(\frac{\text{h}}{2\pi}\big)}{10^{-8}}$
$=\frac{-10^{-34}}{10^{-8}}$
$\Big[\because\frac{\text{h}}{2\pi}\approx10^{-34}\text{J}-\text{s}\Big]$
$=-10^{-42}\text{N}-\text{m}$
The magnitude of the torque is $10^{-42}Nm$.

View full question & answer→

MCQ 161 Mark

In a laboratory experiment on emission from atomic hydrogen in a discharge tube, only a small number of lines are observed whereas a large number of lines are present in the hydrogen spectrum of a star. This is because in a laboratory:

A
The amount of hydrogen taken is much smaller than that present in the star.
✓
The temperature of hydrogen is much smaller than that of the star.
C
The pressure of hydrogen is much smaller than that of the star.
D
The gravitational pull is much smaller than that in the star.

Answer

Correct option: B.

The temperature of hydrogen is much smaller than that of the star.

The number of lines of the hydrogen spectrum depends on the excitation of the hydrogen atom. This is dependent on the heat energy absorbed by the hydrogen atoms. More the temperature of the hydrogen sample, more is the heat energy. The temperature of hydrogen at the star is much more than that can be produced in the laboratory. Hence, less number of lines are observed in the hydrogen spectrum in the laboratory than that in a star.

View full question & answer→

MCQ 171 Mark

A hydrogen atom in ground state absorbs $10.2 eV$ of energy. The orbital angular momentum of the electron is increased by:

✓
$1.05 \times 10^{-34} \mathrm{Js}$
B
$2.11 \times 10^{-34} \mathrm{Js}$
C
$3.16 \times 10^{-34} \mathrm{Js}$
D
$4.22 \times 10^{-34} \mathrm{Js}$

Answer

Correct option: A.

$1.05 \times 10^{-34} \mathrm{Js}$

Let after absorption of energy, the hydrogen atom goes to the $n^{th}$ excited state.
Therefore, the energy absorbed can be written as,
$10.2=13.6\times\Big(\frac{1}{1^2}-\frac{1}{\text{n}^2}\Big)$
$\Rightarrow\frac{10.2}{13.6}=1-\frac{1}{\text{n}^2}$
$\Rightarrow\frac{1}{\text{n}^2}=\frac{13.6-10.2}{13.6}$
$\Rightarrow\frac{1}{\text{n}^2}=\frac{3.4}{13.6}$
$\Rightarrow\text{n}^2=4$
$\Rightarrow\text{n}=2$
The orbital angular momentum of the electron in the $n^{th}$ state is given by,
$\text{L}_\text{n}=\frac{\text{nh}}{2\pi}$
Change in the angular momentum, $\Delta\text{L}=\frac{2\text{h}}{2\pi}-\frac{\text{h}}{2\pi}=\frac{\text{h}}{2\pi}$
$\therefore\Delta\text{T}=1.05\times10^{-34}\text{Js}$

View full question & answer→

MCQ 181 Mark

Which of the following parameters are the same for all hydrogen-like atoms and ions in their ground states?

A
Radius of the orbit.
B
Speed of the electron.
C
Energy of the atom.
✓
Orbital angular momentum of the electron.

Answer

Correct option: D.

Orbital angular momentum of the electron.

According to Bohr's atomic theory, the orbital angular momentum of an electron in a one-electron system is given by,
$\text{L}_\text{n}=\frac{\text{nh}}{2\pi}$
Here,
n = Principal quantum number.
The angular momentum is independent of the atomic number of the one-electron system. Therefore, it is same for all hydrogen-like atoms and ions in their ground states. The other parameters given here are dependent on the atomic number of the hydrogen-like atom or ion taken.

View full question & answer→

Generate Paper Free All Bohr’s Model and Physics of Atom topics