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

Question 1. The increasing order (lowest first) for the values of e/m (charge/mass) for

e,p,n,α
n,p,e,α
n,p,α,e
n,α,p,e

Explanation:-

Answer: Option D. -> n,α,p,e
:
D

\frac{e}{m}

foproduct trackeR
(i) neutron =

\frac{0}{1} = 0

(ii)

α

- Particle

= \frac{2}{4} = 0.5

(iii) Proton

= \frac{1}{1} = 1

(iv) Electron

= \frac{1}{\frac{1}{1837}} = 1837

Question 2. The de-Brogile wavelength of an electron moving in a circular orbit is $λ$ . The minimum radius of orbit is

λπ
λ2π
λ4π
λ3π

Explanation:-

Answer: Option B. -> λ2π
:
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

Explanation:-

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. 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%

Explanation:-

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%.

Question 5. The correct ground state electronic configuration of chromium atom is

[Ar]3d5 4s1
[Ar]3d4 s2
[AR]3d6 4s0
[Ar]4d5 4s1

Explanation:-

Answer: Option A. -> [Ar]3d5 4s1
:
A
3d subshell filled with 5 electrons (half-filled) is more stable than that filled with 4 electrons. 1, 4s electrons jumps into 3d subshell for more sability.

Question 6. Rutherford's alpha particle scattering experiment eventually led to the conclusion that

Mass and energy are related
Electrons occupy space around the nucleus
Neutrons are buried deep in the nucleus
The point of impact with matter can be precisely determined

Explanation:-

Answer: Option B. -> Electrons occupy space around the nucleus
:
B
Electrons in an atom occupy the extra nuclear region.

Question 7. The maximum number of 4f electrons having spin quantum number,

s = \frac{- 1}{2}

is:

14
7
1
3

Explanation:-

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 8. Nitrogen has the electronic configuration

1 s^{2}, 2 s^{2} 2 p_{x}^{1} 2 p_{y}^{1} 2 p_{z}^{1}

and not

1 s^{2}, 2 s^{2} 2 p_{x}^{2} 2 p_{y}^{1} 2 p_{z}^{0}

which is determined by

Aufbau's principle
Pauli's exclusion principle
Hund's rule
Uncertainty principle

Explanation:-

Answer: Option C. -> Hund's rule
:
C
Hund's rule states that pairing of electrons in the orbitals of a subshell (orbitals of equal energy) starts when each of them is singly filled.

Question 9. The spin magnetic momentum of electrons in an ion is 4.84 BM. Its total spin will be

±1
±2
≥ √h4π
±2.5

Explanation:-

Answer: Option B. -> ±2
:
B
<p>

\sqrt{n (n + 2)}

= 4.84 n(n + 2) = 24 n =4 s =

r_{1}

+

r_{2}

+

r_{3}

+

r_{4}

=

\frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2}

= 2 or =

\frac{- 1}{2} \frac{- 1}{2} \frac{- 1}{2} \frac{- 1}{2}

= -2</p>

Question 10. 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 × 1016 s−1
4 × 1016 s−1
8 × 1016 s−1
1 × 1016 s−1

Explanation:-

Answer: Option D. -> 1 × 1016 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}