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Different isotopes of the same element emit light at slightly different wavelengths. A wavelength in the emission spectrum of a hydrogen atom is \(656.45 \mathrm{nm} ;\) for deuterium, the corresponding wavelength is \(656.27 \mathrm{nm}\). (a) What minimum number of slits is required to resolve these two wavelengths in second order? (b) If the grating has 500.00 slits/mm, find the angles and angular separation of these two wavelengths in the second order.

Short Answer

Expert verified
a) The minimum number of slits required to resolve these two wavelengths in the second order is the rounded up value obtained from the first step calculation. b) For the grating has 500.00 slits/mm, the angles of these two wavelengths and their separation can be found using the calculations from step 2 and 3.

Step by step solution

01

Determining the Minimum Number of Slits

Using the formula for the minimum number of slits required to resolve two wavelengths in the second order, which is \(N = \frac{\lambda}{\Delta \lambda} \cdot m\). Here, \(m = 2\) (given in the problem), \(\lambda = 656.45 \mathrm{nm}\) and \(\Delta \lambda = 656.45 \mathrm{nm} - 656.27 \mathrm{nm} = 0.18 \mathrm{nm}\). After substituting these values into the formula, compute the minimum number of slits required.
02

Calculating the Angles

Apply the diffraction grating equation to find the angles of the two wavelengths for the second order, using the formula \(d \sin \theta = m \lambda\), where \(m=2\), \(d = 1 / (500.00 \, \text{slits/mm} \times 10^6 \, \text{mm/m})\), and \(\lambda\) are the wavelengths of hydrogen and deuterium respectively (656.45 nm and 656.27 nm). Substitute these values into the equation and solve for \(\theta\) to find the angles.
03

Finding the Angular Separation

The angular separation of the two wavelengths can be found by subtracting the smaller angle from the larger one.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wavelength Resolution
Understanding wavelength resolution is crucial when analyzing light spectra, particularly when distinguishing between similar wavelengths emitted by different isotopes, such as hydrogen and deuterium. Wavelength resolution refers to the ability of a spectroscope or a diffraction grating to differentiate between two closely spaced wavelengths. The higher the resolution, the closer together two lines can be while still being seen as distinct.

The formula relevant to resolving power is typically expressed as \( R = \frac{\lambda}{\Delta\lambda} \) where \( R \) is the resolving power, \( \lambda \) is the average wavelength of the two lines, and \( \Delta\lambda \) represents their difference in wavelength. In the context of a diffraction grating, as the number of slits (\( N \) increases, so does the resolving power, allowing us to clearly distinguish between finer details in emission spectra. For a grating, the expression for resolving power transforms into \( N = \frac{\lambda}{\Delta\lambda} \cdot m \) with \( m \) denoting the order. This is why, for the given exercise, increasing the number of slits is essential to resolve the near-identical emission lines of hydrogen and deuterium.
Isotopes Emission Spectra
Isotopes of the same element can emit slightly different wavelengths of light due to variations in their nuclear masses. This phenomenon is observable in emission spectra, which are like unique fingerprints for different elements and their isotopes. In the case of hydrogen and deuterium, their emission spectra are almost identical, but with careful observation under high-resolution instruments, slight differences can be detected.

Emission occurs when electrons in the excited state release energy and move to a lower energy level, generating a photon with a specific wavelength. The mass of the nucleus affects the energy levels slightly and thus changes the emitted wavelength. Differentiating these wavelengths not only helps to identify isotopes but also enhances our understanding of atomic structures. For such precision, a high-resolution instrument like a diffraction grating, with many slits, becomes indispensable.
Light Diffraction Angles
The angles at which light is diffracted, a phenomenon governed by the interaction of light with the slit of a diffraction grating, are calculated using the grating equation \( d \sin \theta = m \lambda \). Here \( \theta \) is the diffraction angle, \( d \) is the distance between adjacent slits (also known as grating spacing), \( m \) is the diffraction order, and \( \lambda \) is the wavelength of light.

When a spectrum is produced by a diffraction grating, a variety of angles are observed. Each wavelength diffracts at a unique angle, leading to its position within the spectrum. In the exercise, by applying the grating equation separately for hydrogen and deuterium (with their respective wavelengths), we obtain two distinct angles corresponding to the second-order diffraction (where \( m=2 \) ). The angular separation, which is the difference between these angles, is critical as it indicates whether the grating can distinguish between the two different isotopes' emissions. A larger angular separation typically correlates with easier identification of distinct spectral lines.

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Most popular questions from this chapter

A slit \(0.360 \mathrm{~mm}\) wide is illuminated by parallel rays of light that have a wavelength of \(540 \mathrm{nm}\). The diffraction pattern is observed on a screen that is \(1.20 \mathrm{~m}\) from the slit. The intensity at the center of the central maximum \(\left(\theta=0^{\circ}\right)\) is \(I_{0}\). (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) What is the distance on the screen from the center of the central maximum to the point where the intensity has fallen to \(I_{0} / 2 ?\)

An opaque barrier has an inner membrane and an outer membrane that slide past each other, as shown in Fig. \(\mathbf{P 3 6 . 6 5 .}\) Each membrane includes parallel slits of width \(a\) separated by a distance \(d\). A screen forms a circular arc subtending \(60^{\circ}\) at the fixed midpoint between the slits. A green \(532 \mathrm{nm}\) laser impinges on the slits from the left. The outer membrane moves upward with speed \(v\) while the inner membrane moves downward with the same speed, propelled by nanomotors. At time \(t=0,\) point \(P\) on the outer membrane is adjacent to point \(Q\) on the inner membrane so that the effective aperture width is zero. The aperture is fully closed again at \(t=3.00 \mathrm{~s}\). (a) At \(t=1.00 \mathrm{~s}\), there are 19 evenly spaced bright spots on the screen, each of approximately the same intensity. At the edges of the screen the first diffraction minimum and a two-slit interference maximum coincide. What is the slit distance \(d ?\) (Note: The screen does not encompass the entire diffraction pattern.) (b) What is the speed \(v ?\) (c) What is the maximum aperture width \(a ?\) (d) At a certain time, the outermost spots (the \(m=\pm 9\) spots) disappear. What is that time? (e) At \(t=1.50 \mathrm{~s}\) what is the intensity of the \(m=\pm 1\) spots in terms of the \(m=0\) central spot? (f) What are the angular positions of these spots?

(a) What is the wavelength of light that is deviated in the first order through an angle of \(13.5^{\circ}\) by a transmission grating having 5000 slits \(/ \mathrm{cm} ?\) (b) What is the second-order deviation of this wavelength? Assume normal incidence.

A diffraction grating has 650 slits \(/ \mathrm{mm}\). What is the highest order that contains the entire visible spectrum? (The wavelength range of the visible spectrum is approximately \(380-750 \mathrm{nm}\).)

When laser light of wavelength \(632.8 \mathrm{nm}\) passes through a diffraction grating, the first bright spots occur at \(\pm 17.8^{\circ}\) from the central maximum. (a) What is the line density (in lines/cm) of this grating? (b) How many additional bright spots are there beyond the first bright spots, and at what angles do they occur?

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