Atoms Test

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Atoms Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

The values of $$n_{1}$$ and $$n_{2}$$ for Pfund's series are

A
$$n_{1}=5, n_{2}=6,7$$

B
$$n_{1}=4, n_{2}=5,6,7....$$

C
$$n_{1}=3, n_{2}=4,5,6....$$

D
$$n_{1}=2, n_{2}=3,4,5....$$

Solution

The P'fund series consists of all wavelengths which are emitted when an electron jumps from an outer orbit to the fifth orbit, hence $$n_1 = 5$$ and $$n_2 = 6, 7, 8, ..........., \infty$$. This series lies in infrared region.
Question 2
1 / -0

The maximum possible values of magnetic orbital quantum number ($$m_{l})$$ are

A
$$(2l + 1)$$

B
$$2l $$

C
$$(2l - 1)$$

D
zero

Solution

The principal quantum number n, belongs to the principal shell z$$(K,L,M, ....)$$ which is occupied by an electron and can be any positive integer $$(n = 1, 2, 3, 4, ........)$$. Its location is further narrowed down by the angular momentum quantum number l, corresponding to the subshell and its general shape. The value of l is dependent on the value of $$n (l = 0, 1, 2, .........., n-1)$$. The values of l correspond to specific subshells i.e. l = 0 for s; l = 1 for p; l = 2 for d; l = 3 for f.
The magnetic quantum number($$m_l$$), further divides these subshells into orbitals and gives orientation of these orbitals in space. The number of orbitals can be determined from the angular momentum quantum number l and the maximum possible values of $$m_l$$ can be determined from number of orbitals given by, $$(2l + 1).$$
Question 3
1 / -0

The maximum number of electrons in a shell is

A
$$n^{2}$$

B
$$2n^{2}$$

C
$$2n$$

D
$$n$$

Solution

The number of electrons in any shell is equal to: $$2n^{2}$$
(2, 8 , 8, 16, 32)
Question 4
1 / -0

The fine structure of hydrogen spectrum can be explained by

A
the presence of neutrons in the nucleus.

B
the finite size of nucleus.

C
the orbital angular momentum of electrons.

D
the spin angular momentum of electrons.

Solution

Answer is (D).
The fine structure describes the splitting of the spectral lines of atoms due to electron spin angular momentum.
Question 5
1 / -0

The possible values of orbital quantum numbers are:
[$$n$$ is the principle quantum number]

A
from 0 to ($$n$$-1)

B
from 0 to $$n$$

C
from 0 to ($$n$$+1)

D
all of the above

Solution

The orbital quantum number, l, divides the shells up into smaller groups of subshells called orbitals. The orbital quantum number describes shape of subshells. The principal quantum number(n) determines the possible values of l. For $$n=1$$ i.e. K-shell there is no subshell so, $$l = 0$$ i.e. $$(n-1)$$. Hence, The possible values of orbital quantum numbers are from 0 to $$(n-1)$$.
Question 6
1 / -0

The energy of electron in an excited hydrogen atom is $$-3.4eV$$. Its angular momentum according to Bohr's theory will be

A
$$\displaystyle \dfrac {h}{\pi}$$

B
$$\displaystyle \dfrac {h}{2\pi}$$

C
$$\displaystyle \dfrac {3h}{2\pi}$$

D
$$\displaystyle \dfrac {3}{2\pi h}$$

Solution

Given,
$$E=-3.4eV$$
Energy, $$E=\dfrac{-13.6eV}{n^2}$$
$$n^2=\dfrac{-13.6eV}{-3.4eV}$$
$$n^2=4$$
$$n=2$$
From Bohr's theory, the angular momentum id given by
$$L=\dfrac{nh}{2\pi}$$
$$L=\dfrac{(2)h}{2\pi}$$
$$L=\dfrac{h}{\pi}$$
The correct option is A.
Question 7
1 / -0

The photoelectric effect proves that light consists of

A
Photons

B
Electrons

C
Electromagnetic waves

D
Mechanical waves

Solution

It's been determined experimentally that when light shines on a metal surface, the surface emits electrons. For example, you can start a current in a circuit just by shining a light on a metal plate. we were saying earlier that light is made up of electromagnetic waves, and that the waves carry energy. So if a wave of light hit an electron in one of the atoms in the metal, it might transfer enough energy to knock the electron out of its atom. Light has sometimes been viewed as a particle (photon) rather than a wave. If it's waves, the energy contained in one of those waves should depend only on its amplitude--that is, on the intensity of the light.
So, the answer is option (A).
Question 8
1 / -0

What would be the appropriate quantum number $$n$$ $$A^{0}$$ for circular orbits of radius $$13.2$$ $$A^{0}$$ in hydrogen?

A
$$6$$

B
$$7$$

C
$$5$$

D
$$4$$

Solution

Radius of orbit for quauntum number $$n$$ according to Bohr's Model is
$$r_n=0.529\dfrac{n^2}{Z}A^{\circ}$$
$$\implies 13.2A^{\circ}=0.529n^2A^{\circ}$$
$$\implies n=5$$
Question 9
1 / -0

In which of the following systems, the radius of first orbit ($$n=1$$) is minimum

A
Hydrogen

B
Deuterium atom

C
Singly ionized helium

D
Doubly ionised lithium

Solution

Solving $$\dfrac{mv^2}{r}=\dfrac{1}{4\pi\epsilon_0}\dfrac{e^2}{r^2}$$ and $$mvr=\dfrac{nh}{2\pi}$$ gives,
Radius of nth Bohr orbit is given as - $$r_n=\dfrac{n^2h^2\epsilon_0}{Z\pi me^2}\propto \dfrac{1}{m}$$
Hence it is minimum for lithium since its mass is maximum among the given options.
Question 10
1 / -0

The spin of photon is

A
$$1$$

B
$$ \dfrac { 1 }{ 2 } $$

C
$$ \dfrac { 1 }{ 3 } $$

D
None

Solution

Spin is intrinsic angular momentum and is quantized (as is all angular momentum). Photons are spin-1 particles in contrast to electrons being spin 1/2.
So, the answer is option (A).

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Atoms Test - 30

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Atoms Test - 30

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Question 1

$$n_{1}=5, n_{2}=6,7$$

$$n_{1}=4, n_{2}=5,6,7....$$

$$n_{1}=3, n_{2}=4,5,6....$$

$$n_{1}=2, n_{2}=3,4,5....$$

Solution

Question 2

$$(2l + 1)$$

$$2l $$

$$(2l - 1)$$

zero

Solution

Question 3

$$n^{2}$$

$$2n^{2}$$

$$2n$$

$$n$$

Solution

Question 4

the presence of neutrons in the nucleus.

the finite size of nucleus.

the orbital angular momentum of electrons.

the spin angular momentum of electrons.

Solution

Question 5

from 0 to ($$n$$-1)

from 0 to $$n$$

from 0 to ($$n$$+1)

all of the above

Solution

Question 6

$$\displaystyle \dfrac {h}{\pi}$$

$$\displaystyle \dfrac {h}{2\pi}$$

$$\displaystyle \dfrac {3h}{2\pi}$$

$$\displaystyle \dfrac {3}{2\pi h}$$

Solution

Question 7

Photons

Electrons

Electromagnetic waves

Mechanical waves

Solution

Question 8

$$6$$

$$7$$

$$5$$

$$4$$

Solution

Question 9

Hydrogen

Deuterium atom

Singly ionized helium

Doubly ionised lithium

Solution

Question 10

$$1$$

$$ \dfrac { 1 }{ 2 } $$

$$ \dfrac { 1 }{ 3 } $$

None

Solution

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