12th Grade > Chemistry
ATOMIC STRUCTURE MCQs
Total Questions : 15
| Page 1 of 2 pages
Answer: Option D. -> n,α,p,e
:
D
em foproduct trackeR
(i) neutron = 01=0
(ii) α - Particle=24=0.5
(iii) Proton =11=1
(iv) Electron =111837=1837
:
D
em foproduct trackeR
(i) neutron = 01=0
(ii) α - Particle=24=0.5
(iii) Proton =11=1
(iv) Electron =111837=1837
Answer: Option B. -> λ2π
:
B
We know 2πr=nλ
From minimum radius n=1
2πrmin=λ
rmin=λ2π
:
B
We know 2πr=nλ
From minimum radius n=1
2πrmin=λ
rmin=λ2π
Answer: Option C. -> N - shell
:
C
nh2π=4.2178×10−34
n = 2π×4.2178×10−34h
=2×3.14×4.2178×10−346.625×10−34
n = 4
N-shell
:
C
nh2π=4.2178×10−34
n = 2π×4.2178×10−34h
=2×3.14×4.2178×10−346.625×10−34
n = 4
N-shell
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%.
:
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%.
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.
:
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.
Answer: Option B. -> Electrons occupy space around the nucleus
:
B
Electrons in an atom occupy the extra nuclear region.
:
B
Electrons in an atom occupy the extra nuclear region.
Answer: Option B. -> 7
:
B
Maximum number of electrons present in 4f orbitals are 14. Half of the electrons will have +12 spin and rest half will have −12 spin
:
B
Maximum number of electrons present in 4f orbitals are 14. Half of the electrons will have +12 spin and rest half will have −12 spin
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.
:
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.
Answer: Option B. -> ±2
:
B
<p>√n(n+2) = 4.84 n(n + 2) = 24 n =4 s = r1 + r2 + r3 + r4 =12+12+12+12= 2 or = −12−12−12−12 = -2</p>
:
B
<p>√n(n+2) = 4.84 n(n + 2) = 24 n =4 s = r1 + r2 + r3 + r4 =12+12+12+12= 2 or = −12−12−12−12 = -2</p>
Question 10. When a certain metal was irradiated with light of frequency 4.0 × 1016 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 × 1016s−1. Calculate the critical frequency (ν0) of the metal.
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,
KE=hν−hν0=h(ν−ν0)
KE1=h(ν1−ν0) ...(i)
KE2=h(ν2−ν0) ...(ii)
Dividing equation (ii) by (i), we get
KE2KE1=h(ν2−ν0)h(ν1−ν0)=(ν2−ν0)(ν1−ν0)
But given that
KE2KE1=3
∴3=(ν2−ν0)(ν1−ν0)⇒3(ν1−ν0)=ν2−ν0
⇒3ν1−ν2=3ν0−ν0=2ν0
⇒3×2.0×1016−4×1016=2ν0
⇒ν0=2×10162=1×1016s−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,
KE=hν−hν0=h(ν−ν0)
KE1=h(ν1−ν0) ...(i)
KE2=h(ν2−ν0) ...(ii)
Dividing equation (ii) by (i), we get
KE2KE1=h(ν2−ν0)h(ν1−ν0)=(ν2−ν0)(ν1−ν0)
But given that
KE2KE1=3
∴3=(ν2−ν0)(ν1−ν0)⇒3(ν1−ν0)=ν2−ν0
⇒3ν1−ν2=3ν0−ν0=2ν0
⇒3×2.0×1016−4×1016=2ν0
⇒ν0=2×10162=1×1016s−1