11th And 12th > Physics
ATOMS MCQs
Total Questions : 30
| Page 1 of 3 pages
Answer: Option D. ->
n2
:
D
Bohr radius r=∑0n2h2πZme2;∴r∞n2
:
D
Bohr radius r=∑0n2h2πZme2;∴r∞n2
Answer: Option A. ->
10.2 eV
:
A
The energy in the first orbit is - 13.6eV
n=2 --------------- E2=−13.6(2)2=−3.4 eV
E1→2=−3.4−(13.6)=+10.2 eV
:
A
The energy in the first orbit is - 13.6eV
n=2 --------------- E2=−13.6(2)2=−3.4 eV
E1→2=−3.4−(13.6)=+10.2 eV
Answer: Option D. ->
double -ionized lithium
:
D
1λ=RZ2(112−122)
For double-ionised lithium the value of Z is maximum.
:
D
1λ=RZ2(112−122)
For double-ionised lithium the value of Z is maximum.
Answer: Option C. ->
6
:
C
Using 1λ=R[1n21−1n22]⇒1975×10−10=1.097×107(112−1n2)⇒n=4
Now number of spectral lines N=n(n−1)2=4(4−1)2=6
:
C
Using 1λ=R[1n21−1n22]⇒1975×10−10=1.097×107(112−1n2)⇒n=4
Now number of spectral lines N=n(n−1)2=4(4−1)2=6
Answer: Option C. ->
n = 4
:
C
Let electron absorbing the photon energy reaches to the excited state n. Then using energy conservation
⇒−13.6n2=−13.6+12.4⇒−13.6n2=−1.2⇒n2=13.61.2=12⇒n=3.46≃4
:
C
Let electron absorbing the photon energy reaches to the excited state n. Then using energy conservation
⇒−13.6n2=−13.6+12.4⇒−13.6n2=−1.2⇒n2=13.61.2=12⇒n=3.46≃4
Answer: Option D. ->
81,588 cm−1
:
D
Using 1λ=¯v=RZ2(1n21−1n22)⇒¯v∞Z2⇒¯v2¯v1=(Z2Z1)2=(Z1)2=4⇒¯v2=¯v×4=81588 cm−1
:
D
Using 1λ=¯v=RZ2(1n21−1n22)⇒¯v∞Z2⇒¯v2¯v1=(Z2Z1)2=(Z1)2=4⇒¯v2=¯v×4=81588 cm−1
Answer: Option D. ->
18
:
D
Time period T∝n3Z2
For a given atom (Z = constant) So T∝n3 ......... (i) and radius R∝n2 ......... (ii)
∴ From equation (i) and (ii) T∝R3/2⇒T1T2=(R1R2)3/2=(R4R)3/2=18
:
D
Time period T∝n3Z2
For a given atom (Z = constant) So T∝n3 ......... (i) and radius R∝n2 ......... (ii)
∴ From equation (i) and (ii) T∝R3/2⇒T1T2=(R1R2)3/2=(R4R)3/2=18
Answer: Option A. ->
E
:
A
Ionisation energy of atom in nth state EnαZ2n2
For hydrogen atom in ground state (n = 1) and Z = 1 ⇒E=E0 ....... (i)
For Li++ atom in 2nd excited state n = 3 and Z = 3, hence E′=E032×32=E0 ........ (ii)
From equation (i) and (ii) E′=E
:
A
Ionisation energy of atom in nth state EnαZ2n2
For hydrogen atom in ground state (n = 1) and Z = 1 ⇒E=E0 ....... (i)
For Li++ atom in 2nd excited state n = 3 and Z = 3, hence E′=E032×32=E0 ........ (ii)
From equation (i) and (ii) E′=E
Answer: Option C. ->
2 : 1
:
C
Tn∝n3 n∝T13
n1n2=[T1T2]13=[81]13
n1:n2=2:1
:
C
Tn∝n3 n∝T13
n1n2=[T1T2]13=[81]13
n1:n2=2:1
Answer: Option A. ->
3.3×1015HZ
:
A
c=νλ ; ν=cλ=3×108911×10−10
ν=3.3×1015HZ
:
A
c=νλ ; ν=cλ=3×108911×10−10
ν=3.3×1015HZ