Question 53 of 62advanced🔧 ApplyNumerical3 marks

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.

Correct Answer

The wavelength of the emission transition is 434 nm. The series is the Balmer series and the region of the spectrum is the visible region.

Exercise: EXERCISES | Q: 2.56 | (Page 41)
For More Understanding

Explanation

The ratio of the initial and final orbit radii is 1322.5 pm/211.6 pm=6.25=25/41322.5 \text{ pm} / 211.6 \text{ pm} = 6.25 = 25/4. Since the radius of an orbit is proportional to the square of its principal quantum number (n2n^2), the ratio ni2/nf2=25/4n_i^2 / n_f^2 = 25/4, which gives ni=5n_i = 5 and nf=2n_f = 2. This corresponds exactly to the transition calculated in Problem 2.10 of the provided context, yielding a wavelength of 434 nm in the visible region of the Balmer series.

Solution Steps

  1. Step 1: Calculate the ratio of the radii Convert the initial radius to pm: 1.3225 nm=1322.5 pm1.3225 \text{ nm} = 1322.5 \text{ pm}. Ratio of radii = 1322.5 pm211.6 pm=6.25=254\frac{1322.5 \text{ pm}}{211.6 \text{ pm}} = 6.25 = \frac{25}{4}

  2. Step 2: Determine the initial and final orbits Since the radius of an orbit is proportional to n2n^2, we have ni2nf2=254\frac{n_i^2}{n_f^2} = \frac{25}{4}. This gives ni=5n_i = 5 and nf=2n_f = 2.

  3. Step 3: Calculate the energy change (ΔE\Delta E) Using equation (2.17) for ni=5n_i = 5 and nf=2n_f = 2:

    ΔE=2.18×10−18 J[152−122]=−4.58×10−19 J\Delta E = 2.18 \times 10^{-18} \text{ J} \left[ \frac{1}{5^2} - \frac{1}{2^2} \right] = -4.58 \times 10^{-19} \text{ J}

  4. Step 4: Calculate the frequency (vv)

    v=ΔEh=4.58×10−19 J6.626×10−34 J s=6.91×1014 Hzv = \frac{\Delta E}{h} = \frac{4.58 \times 10^{-19} \text{ J}}{6.626 \times 10^{-34} \text{ J s}} = 6.91 \times 10^{14} \text{ Hz}

  5. Step 5: Calculate the wavelength (λ\lambda)

    λ=cv=3.0×108 m s−16.91×1014 Hz=434 nm\lambda = \frac{c}{v} = \frac{3.0 \times 10^8 \text{ m s}^{-1}}{6.91 \times 10^{14} \text{ Hz}} = 434 \text{ nm}

  6. Step 6: Identify the series and spectrum region The transition from n=5n = 5 to n=2n = 2 belongs to the Balmer series, which lies in the visible region of the spectrum.

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