Question

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 1.285 nm. Find the region of the spectrum. 

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Answer 1

Answer :

• Value of n = 5.
• Spectrum lies in the infrared region.

Step-by-step explanation :

Given that,

• Frequency, v = 3.29 × 10¹⁵ [1/3² - 1/n²] Hz.
• Wavelength, λ = 1285 nm = 1285 × 10⁻⁹ m.

Also,

• Speed of light, c = 3 × 10⁸ ms⁻¹

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We know, $\rm{v=\dfrac{c}{\lambda}}$

Putting the values, we get,

$:\implies\rm{3.29\times{}10^{15}\left[\dfrac{1}{3^2}-\dfrac{1}{n^2}\right]=\dfrac{3\times{}10^8\:ms^{-1}}{1285\times{}10^{-9}\:m}}$

$:\implies\rm{\dfrac{1}{n^2}=\dfrac{1}{9}-\dfrac{3\times{}10^8}{1285\times{}10^{-9}}\times{}\dfrac{1}{3.29\times{}10^{15}}}$

$:\implies\rm{\dfrac{1}{n^2}=0.111-0.071=0.04=\dfrac{1}{25}}$

$:\implies\rm{n^2=25}$

$:\implies\rm{n=\sqrt{25}=5}$

Hence, the radiation corresponding to 1285 nm lies in the infrared region.

Answer 2

Answer:

A/C to question,

v = 3.29 × 10^15 ( 1/3² - 1/n²)Hz

we know,

frequency = speed of light/wavelength

given , here

wavelength = 1285 nm = 1.285 × 10^-6 m

speed of light = 3 × 10^8 m/s

frequency ( v ) = 3 × 10^8/1.285 × 10^-6 Hz

3.29 × 10^15 ( 1/3² - 1/n²) Hz = 3 × 10^8/1.285× 10^-6 Hz

( 1/9 - 1/n²) = 3 × 10^8/1.285 × 10^-6 × 3.29 × 10^15

0.1111 - 1/n² = 0.0709

1/n² = 0.1111 - 0.0709 = 0.0402 ≈ 0.04

1/n² = 1/25

1/n² = 1/5²

n = 5

the electrons jumps from n = 5 to n = 3 .e.g., the transition occurs in paschen series and lies infrared region.

Hence, the radiation 1285 nm lies in the infrared region.