Question 52 of 62intermediate🔧 ApplyNumerical3 marks

Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represeted as v = 3.29×103.29 \times 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.

Correct Answer

(n) 5

(region) Infrared

Exercise: EXERCISES | Q: 2.55 | (Page 41)
For More Understanding

Explanation

The frequency of the emitted photon is calculated using the given wavelength and the speed of light. By substituting this frequency into the provided Paschen series frequency equation, we can solve for the initial orbit n. Since the transition belongs to the Paschen series, it falls in the infrared region of the spectrum.

Solution Steps

  1. Step 1: Calculate the frequency (vv) of the transition Given wavelength, λ=1285 nm=1285×10−9 m\lambda = 1285 \text{ nm} = 1285 \times 10^{-9} \text{ m} Speed of light, c=3.0×108 m s−1c = 3.0 \times 10^8 \text{ m s}^{-1}

    v=cλ=3.0×108 m s−11285×10−9 m=2.33×1014 Hzv = \frac{c}{\lambda} = \frac{3.0 \times 10^8 \text{ m s}^{-1}}{1285 \times 10^{-9} \text{ m}} = 2.33 \times 10^{14} \text{ Hz}

  2. Step 2: Substitute vv into the given Paschen series equation

    v=3.29×1015 Hz[132−1n2]v = 3.29 \times 10^{15} \text{ Hz} \left[ \frac{1}{3^2} - \frac{1}{n^2} \right] 2.33×1014=3.29×1015[19−1n2]2.33 \times 10^{14} = 3.29 \times 10^{15} \left[ \frac{1}{9} - \frac{1}{n^2} \right]

  3. Step 3: Solve for nn

    2.33×10143.29×1015=19−1n2\frac{2.33 \times 10^{14}}{3.29 \times 10^{15}} = \frac{1}{9} - \frac{1}{n^2} 0.0708=0.1111−1n20.0708 = 0.1111 - \frac{1}{n^2} 1n2=0.1111−0.0708=0.0403\frac{1}{n^2} = 0.1111 - 0.0708 = 0.0403 n2=10.0403=24.81n^2 = \frac{1}{0.0403} = 24.81 n=24.81≈4.98n = \sqrt{24.81} \approx 4.98

    Since nn must be an integer, n=5n = 5.

  4. Step 4: Determine the region of the spectrum As given in Table 2.3, the Paschen series (where transitions end at n1=3n_1 = 3) lies in the Infrared region of the spectrum.

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