Test Index

NCERT Class XI Chemistry Structure of Atom Solutions

© examsnet.com
Question : 55
Total: 67
Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as υ = 3.29 × 1015 (Hz) [
1
32
−
1
n2
]. Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.
Solution:  
Since l = 1285 nm = 1285 × 10–9m [Since 1 nm = 10–9 m]
∴ v =
c
λ
 =
3×108
1285×10−9
 = 2.33 × 1014sec−1
According to question, v = 3.29 × 1015(
1
32
−
1
n2
)
∴ 2.33 × 1014 = 3.29 × 1015(
1
9
−
1
n2
)
∴
2.33×1014
3.29×1015
 =
1
9
−
1
n2
 or 0.0709 =
1
9
−
1
n2
or
1
n2
 =
1
9
 - 0.0708 or
1
n2
 =
1−0.64
9
 or
1
n2
 =
0.36
9
∴ n2 =
9
0.36
 =
900
36
 = 25 ⇒ n = √25 = 5
l = 1.285 × 10–6m i.e., of order 10–6m which lies in the infrared region.
© examsnet.com
Go to Question:
Prev Question
Next Question