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.

<p>Radius of orbit of H like species = (0.529 / Z) n²Å = (52.9 / Z) n²pm</p><p>r₁ = 1.3225 nm = 1322.5 pm = (52.9 / Z) n₁²</p><p>r₂ = 211.6 pm = 211.6 pm = (52.9 / Z) n₂²</p><p>∴ r₁/ r₂= 1322.5 / 211.6</p><p>=&gt;n₁²/n₂²= 6.25</p><p>=&gt; n₁/n₂= (6.25)^1/2 = 2.5</p><p>=&gt; n₁= 2.5 n₂</p><p>=&gt; 10 n₁= 25 n₂</p><p>=&gt; 2 n₁= 5 n₂</p><p>If n₁= 2, then n₂= 5. That means transition occurs from 5^th orbit to 2^nd orbit. This means that the transition belongs to Balmer series.</p><p>Now, wave number? = (1.097 x 10⁷ m^-1) x (1/2² – 1/5²) = 1.097 x 10⁷ x 21/100 m^-1= 23.037 x 10⁵ m^-1</p><p>λ = 1/? = 1/ 23.037 x 10⁵ m^-1</p><p>= 434 x 10^-9m = 434 nm</p><p>This transition belongs to visible region of the spectrum of light.</p>

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