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Identifying Isotones hy Snectra Different isotones of the same element emit light at slightly different wavelengths. A wavelength in the emission spectrum of a hydrogen atom is 656.45 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) 1824 slits; (b) angles: 41.21°, 41.18°; separation: 0.03°.

Step by step solution

01

Determine the Resolving Power Requirement

The resolving power \( R \) of a grating is given by \( R = \frac{m \cdot N}{\Delta \lambda} = \frac{\lambda}{\Delta \lambda} \), where \( m \) is the order of the spectrum, \( N \) is the number of slits, and \( \Delta \lambda \) is the difference in wavelength that needs to be resolved. First, calculate the change in wavelength, \( \Delta \lambda = 656.45 \text{ nm} - 656.27 \text{ nm} = 0.18 \text{ nm} \).
02

Calculate Minimum Number of Slits

For second-order diffraction, \( m = 2 \). Substitute the known values into the equation:\[ N = \frac{\lambda}{m \cdot \Delta \lambda} = \frac{656.36 \text{ nm}}{2 \cdot 0.18 \text{ nm}} \approx 1824 \]Thus, the minimum number of slits required is 1824.
03

Calculate Diffraction Angle for Each Wavelength

The angle \( \theta \) for a grating is given by \( d \cdot \sin \theta = m \cdot \lambda \), where \( d = \frac{1}{500 \times 10^3} \text{ mm} \) is the distance between slits. Calculate for each wavelength:For \( \lambda_1 = 656.45 \text{ nm}\):\[ \sin \theta_1 = \frac{2 \times 656.45 \times 10^{-9} \text{ m}}{2 \times 10^{-6} \text{ m}} \approx 0.65645 \]\[ \theta_1 \approx \arcsin(0.65645) \approx 41.21^\circ \]For \( \lambda_2 = 656.27 \text{ nm}\):\[ \sin \theta_2 = \frac{2 \times 656.27 \times 10^{-9} \text{ m}}{2 \times 10^{-6} \text{ m}} \approx 0.65627 \]\[ \theta_2 \approx \arcsin(0.65627) \approx 41.18^\circ \]
04

Find the Angular Separation

The angular separation \( \Delta \theta \) is given by:\[ \Delta \theta = \theta_1 - \theta_2 \approx 41.21^\circ - 41.18^\circ = 0.03^\circ \]

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Key Concepts

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

Resolving Power
The concept of resolving power is crucial in understanding how effectively a diffraction grating can distinguish between two closely spaced wavelengths in an emission spectrum. Resolving power, denoted as \( R \), is defined by the relation \( R = \frac{m \cdot N}{\Delta \lambda} = \frac{\lambda}{\Delta \lambda} \). Here, \( m \) represents the order of the diffraction spectrum, \( N \) is the number of slits in the grating, and \( \Delta \lambda \) is the wavelength difference being measured. To calculate the resolving power needed, you first identify the smallest wavelength difference you wish to resolve, which in this context is between hydrogen and deuterium emissions. For successful resolution at the second-order spectrum, it's essential to have a grating with a minimum number of slits. This ensures that light of slightly varying wavelengths can be separated, thus showing distinct spectral lines. A greater number of slits increases the resolving power, enhancing the ability to distinguish between very close spectral lines. Understanding this property is vital for applications such as spectroscopy, where precise resolution of spectral lines can lead to insights into the material composition.
Diffraction Angle Calculations
Calculating the diffraction angle provides insights into how light, as it passes through the grating, bends at specific angles based on its wavelength. The formula \( d \cdot \sin \theta = m \cdot \lambda \) calculates this angle, \( \theta \), where \( d \) represents the distance between the grating slits (also known as spacing). Given the specific wavelengths and a grating with 500 slits per millimeter, the slit spacing \( d \) becomes \( \frac{1}{500 \times 10^3} \) meters, converting from millimeters. For each wavelength, you substitute its value into the equation to find its diffraction angle. For example, the hydrogen wavelength gives a slightly larger angle than deuterium due to its longer wavelength. These calculations are significant, as they allow one to determine where exactly on a detector or screen the light will form bright spots, aiding in the precise investigation of the emission spectra.
Wavelengths in Emission Spectra
Emission spectra allow us to understand the different energies emitted by atoms or molecules when they transition from a higher energy state to a lower one. Each element or isotope can emit light at characteristic wavelengths, forming a spectrum that can be seen as a series of lines when viewed through a diffraction grating. For isotones such as hydrogen and deuterium, even slight differences in mass lead to slightly different wavelengths, as observed with 656.45 nm and 656.27 nm, respectively. These tiny variations are pivotal in scientific research, often requiring precise instruments like diffraction gratings with high resolving power to decode. Understanding these differences helps researchers identify elements and isotopes in various samples, ranging from stellar spectra in astrophysics to intricate chemical analyses. By revealing unique spectral signatures, emission spectra serve as fingerprints for substances, enabling new discoveries in both scientific and industrial applications.

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Most popular questions from this chapter

An interference pattern is produced by four parallel and equally spaced, narrow slits. By drawing appropriate phasor diagrams, show that there is an interference minimum when the phase difference \(\phi\) from adjacent slits is (a) \(\pi / 2 ;\) (b) \(\pi ;(\mathrm{c}) 3 \pi / 2\) . In each case, for which pairs of slits is there totally destructive interference?

The light from an iron arc includes many different wavelengths. Two of these are at \(\lambda=587.9782 \mathrm{nm}\) and \(\lambda=\) 587.8002 \(\mathrm{nm} .\) You wish to resolve these spectral lines in first order using a grating 1.20 \(\mathrm{cm}\) in length. What minimum number of slits per centimeter must the grating have?

Two satellites at an altitude of 1200 \(\mathrm{km}\) are separated by 28 km. If they broadcast \(3.6-\mathrm{cm}\) microwaves, what minimum receiving-dish diameter is needed to resolve (by Rayleigh's criterion) the two transmissions?

Monochromatic light from a distant source is incident on a slit 0.750 \(\mathrm{mm}\) wide. On a screen 2.00 \(\mathrm{m}\) away, the distance from the central maximum of the diffraction pattern to the first minimum is measured to be 1.35 \(\mathrm{mm}\) . Calculate the wavelength of the light.

Monochromatic light of wavelength 486 \(\mathrm{nm}\) from a distant source passes through a slit that is 0.0290 \(\mathrm{mm}\) wide. In the resulting diffraction pattern, the intensity at the center of the central maximum \(\left(\theta=0^{\circ}\right)\) is \(4.00 \times 10^{-5} \mathrm{W} / \mathrm{m}^{2} .\) What is the intensity at a point on the screen that corresponds to \(\theta=1.20^{\circ} .\)

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