Atomic Structure(11th And 12th > Chemistry ) Questions and answers for exam Preparation

11th And 12th > Chemistry

ATOMIC STRUCTURE MCQs

Total Questions : 15 | Page 1 of 2 pages

Uncertainity in position of a hypothetical subatomic particle is $1 Å$ and uncertainty in velocity is $\frac{3.3}{4 π} \times 10^{5} m / s$ then the mass of the particle is approximately $[h = 6.6 \times 10^{- 34} J s]$

$2 \times 10^{- 28} k g$
$2 \times 10^{- 27} k g$
$2 \times 10^{- 29} k g$
$2 \times 10^{- 26} k g$

Discuss Question

Answer: Option C. ->

2 \times 10^{- 29} k g

:
C

$Δ x \times m \times Δ v \geq \frac{h}{4 π}$

$1 \times 10^{- 10} \times m \times \frac{3.3}{4 π} \times 10^{5} \geq \frac{6.6 \times 10^{- 34}}{4 \times π}$

$M = 2 \times 10^{- 29} k g$

Question 2.

The de-Brogile wavelength of an electron moving in a circular orbit is $λ$ . The minimum radius of orbit is

$\frac{λ}{π}$
$\frac{λ}{2 π}$
$\frac{λ}{4 π}$
$\frac{λ}{3 π}$

Discuss Question

Answer: Option B. ->

\frac{λ}{2 π}

:
B

We know $2 π r = n λ$

From minimum radius $n = 1$

$2 π r_{m i n} = λ$

$r_{m i n} = \frac{λ}{2 π}$

Question 3.

The electron belongs to the angular momentum of an $e^{-}$ in a Bohr's orbit of H atom is $4.2178 \times 10^{- 34} K g . m^{2} . S e c^{- 1}$ .

K - shell
L - shell
N - shell
M - shell

Discuss Question

Answer: Option C. -> N - shell
:
C

$\frac{n h}{2 π}$ = $4.2178 \times 10^{- 34}$

$n$ = $\frac{2 π \times 4.2178 \times 10^{- 34}}{h}$

= $\frac{2 \times 3.14 \times 4.2178 \times 10^{- 34}}{6.625 \times 10 - 34}$

$n$ = $4$

N-shell

Question 4.

When a certain metal was irradiated with light of frequency $4.0 \times 10^{16} s^{- 1}$ , the photoelectrons emitted had three times the kinetic energy as the kinetic energy of photoelectrons emitted when the metal was irradicated with light of frequency $2.0 \times 10^{16} s^{- 1}$ . Calculate the critical frequency ( $ν_{0}$ ) of the metal.

$2 \times 10^{16} s^{- 1}$
$4 \times 10^{16} s^{- 1}$
$8 \times 10^{16} s^{- 1}$
$1 \times 10^{16} s^{- 1}$

Discuss Question

Answer: Option D. ->

1 \times 10^{16} s^{- 1}

:
D

We know that, the incident photons should have a minimum frequency called threshold frequency and a certain amount of energy required called work function. Therefore,

$K E = h ν - h ν_{0} = h (ν - ν_{0})$

$K E_{1} = h (ν_{1} - ν_{0})$ ...(i)

$K E_{2} = h (ν_{2} - ν_{0})$ ...(ii)

Dividing equation (ii) by (i), we get

$\frac{K E_{2}}{K E_{1}} = \frac{h (ν_{2} - ν_{0})}{h (ν_{1} - ν_{0})} = \frac{(ν_{2} - ν_{0})}{(ν_{1} - ν_{0})}$

But given that

$\frac{K E_{2}}{K E_{1}} = 3$

$∴ 3 = \frac{(ν_{2} - ν_{0})}{(ν_{1} - ν_{0})} \Rightarrow 3 (ν_{1} - ν_{0}) = ν_{2} - ν_{0}$

$\Rightarrow 3 ν_{1} - ν_{2} = 3 ν_{0} - ν_{0} = 2 ν_{0}$

$\Rightarrow 3 \times 2.0 \times 10^{16} - 4 \times 10^{16} = 2 ν_{0}$

$\Rightarrow ν_{0} = \frac{2 \times 10^{16}}{2} = 1 \times 10^{16} s^{- 1}$

Question 5.

The spin magnetic momentum of electrons in an ion is 4.84 BM. Its total spin will be

$\pm 1$
$\pm 2$
$\geq$ $\sqrt{\frac{h}{4 π}}$
$\pm 2.5$

Discuss Question

Answer: Option B. ->

\pm 2

:
B

Question 6.

Which of the following sets of quantum numbers is correct for an electron in 4f orbital?

$n = 4, 1 = 3, m = + 1, s = + \frac{1}{2}$
$n = 4, 1 = 4, m = - 4, s = - \frac{1}{2}$
$n = 4, 1 = 3, m = + 4, s = + \frac{1}{2}$
$n = 3, 1 = 2, m = - 2, s = + \frac{1}{2}$

Discuss Question

Answer: Option A. ->

n = 4, 1 = 3, m = + 1, s = + \frac{1}{2}

:
A
The possible quantum numbers for 4f electron are
n=4, l=3, m=-3, -2, -1, 0, 1, 2, 3 and

s = \pm \frac{1}{2}

Question 7.

The correct set of quantum numbers for the unpaired electron of chlorine atom (Z=17) is:

n = 2, l = 1, m = 0
n = 2, l = 1, m = 1
n = 3, l = 1, m = 1
n = 3, l = 0, m = 0

Discuss Question

Answer: Option C. -> n = 3, l = 1, m = 1
:
C
atomic number of Cl=17 , electronic configuration =

1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5}

For 3p orbital n=3, l=1 m=-1,0,+1

Question 8.

Which of the following pair is isodiapheres?

$C_{6}^{14} a n d N a_{11}^{23}$
$M g_{12}^{24} a n d N a_{11}^{23}$
$H e_{2}^{4} a n d O_{8}^{16}$
$C_{6}^{12} a n d N_{7}^{15}$

Discuss Question

Answer: Option C. ->

H e_{2}^{4} a n d O_{8}^{16}

:
C
Isodiaphers have same value of A – (2Z)

Question 9.

The maximum number of 4f electrons having spin quantum number, $s = \frac{- 1}{2}$ is:

Discuss Question

Answer: Option B. -> 7
:
B
Maximum number of electrons present in 4f orbitals are 14. Half of the electrons will have

+ \frac{1}{2}

spin and rest half will have

- \frac{1}{2}

spin

Question 10.

Atoms consists of protons, neutrons and electrons. If the mass of neutrons and electrons were made half and two times respectively to their actual masses, then the atomic mass of $_{6} C^{12}$

Will remain approximately the same
Will become approximately two times
Will remain approximately half
Will be reduced by 25%

Discuss Question

Answer: Option D. -> Will be reduced by 25%
:
D

No change by doubling mass of electrons however by reducing mass of neutron to half total atomic mass becomes 6+3 instead of 6+6. Thus reduced by 25%.