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Class 11
Chemistry
Structure of Atom
calculate the wavelength for the emission transiti...

Question:
Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.

Answer:

The radius of the n^th orbit of hydrogen-like particles = 0.529n²/Z Å

Now r₁ = 1.3225 nm or 1322.5 pm = 52.9n₁²

And

r₂ = 211.6pm = 52.9n₂²/Z

Therefore r₁/r₂ = 1322.5 / 211.6 = n₁²/n₂²

or n₁²/n₂² = 6.25

or n₁/n₂ = 2.5

therefore n₂ = 2 , n₁ = 5.

Thus the transition is from 5^th orbit to 2^nd orbit. It belongs Balmer series

Wave number for the transition is given by,

1.097 × 10⁷ m^–1 (1/2²-1/5²)

=1.097 x 10⁷m^-1 (21/100)

= 2.303 × 10⁶ m^–1

Wavelength (λ) associated with the emission transition is given by,

= 0.434 ×10^–6 m

λ = 434 nm

the region is visible region

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Question:
Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.

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Question: Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.

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Question:
Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.