Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v = 3.29 × 1015 (Hz) [1/32 – 1/n2] - Chemistry

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Numerical

Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v = 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.

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Solution 1

Wavelength of transition = 1285 nm

= 1285 × 10–9 m (Given)

`v = 3.29 xx 10^(15)(1/3^2 - 1/"n"^2)` (Given)

Since `"v" = "c"/lambda`

`= (3.0 xx 10^8)/(1285xx10^(-9) " m")`

ν = 2.33 × 1014 s–1

Substituting the value of ν in the given expression,

`3.29 xx 10^(15)(1/9 - 1/"n"^2) = 2.33 xx 10^14`

`1/9 - 1/"n"^2 = (2.33xx10^14)/(3.29xx10^15)`

`1/9 - 0.7082 xx 10^(-1) = 1/"n"^2 `

`=> 1/"n"^2 = 1.1 xx 10^(-1) - 0.7082 xx 10^(-1)`

`1/"n"^2 = 4.029 xx 10^(-2)`

n = `sqrt(1/(4.029 xx 10^(-2)))`

n = 4.98

n ≈ 5

Hence, for the transition to be observed at 1285 nm, n = 5.

The spectrum lies in the infra-red region.

Solution 2

ν = c/λ = 3.0×10ms-1/1285×10-9 m = 3.29×1015 (1/3- 1/n2)

⇒ 1/n= 1/9 - (3.0×10ms-1/1285×10-9 m)×(1/3.29×1015) = 0.111-0.071 = 0.04 = 1/25

⇒ n= 25

⇒ n = 5

The radiation corresponding to 1285 nm lies in the infrared region.

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Chapter 2: Structure of Atom - EXERCISES [Page 72]

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NCERT Chemistry Part 1 and 2 Class 11
Chapter 2 Structure of Atom
EXERCISES | Q 2.55 | Page 72
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