Atoms(11th And 12th > Physics ) Questions and Answers
Explanation:-
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
Explanation:-
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
Explanation:-
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
Explanation:-
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
Explanation:-
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
Explanation:-
Answer: Option C. -> 2 : 1:
C
Tn∝n3 n∝T13
n1n2=[T1T2]13=[81]13
n1:n2=2:1