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NCERT Class XI Chemistry Structure of Atom Solutions

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Question : 55 of 67
Marks: +1, -0
Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as υ = 3.29 × 101510^{15} (Hz) [132−1n2]\left[ \frac{1}{3^2} - \frac{1}{n^2} \right]. 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 × 10−9 m10^{-9} \text{ m} [Since 1 nm = 10−910^{-9} m]
∴ v = cλ\frac{c}{\lambda} = 3×1081285×10−9\frac{3 \times 10^8}{1285 \times 10^{-9}} = 2.33 × 1014 sec−110^{14} \text{ sec}^{-1}
According to question, v = 3.29 × 1015(132−1n2)10^{15} \left( \frac{1}{3^2} - \frac{1}{n^2} \right)
∴ 2.33 × 101410^{14} = 3.29 × 1015(19−1n2)10^{15} \left( \frac{1}{9} - \frac{1}{n^2} \right)
∴ 2.33×10143.29×1015\frac{2.33 \times 10^{14}}{3.29 \times 10^{15}} = 19−1n2\frac{1}{9} - \frac{1}{n^2} or 0.0709 = 19−1n2\frac{1}{9} - \frac{1}{n^2}
or 1n2\frac{1}{n^2} = 19\frac{1}{9} - 0.0708 or 1n2\frac{1}{n^2} = 1−0.649\frac{1-0.64}{9} or 1n2\frac{1}{n^2} = 0.369\frac{0.36}{9}
∴ n2n^2 = 90.36\frac{9}{0.36} = 90036\frac{900}{36} = 25 ⇒ n = 25\sqrt{25} = 5
l = 1.285 × 10−6 m10^{-6} \text{ m} i.e., of order 10−6 m10^{-6} \text{ m} which lies in the infrared region.
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