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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 nm; for deuterium, the corresponding wavelength is 656.27 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) Minimum slits required: 329. (b) Angles: \(41.15^\circ\) and \(41.13^\circ\); Angular separation: \(0.02^\circ\).

Step by step solution

01

Understanding the Problem

We are given two wavelengths: one for hydrogen, 656.45 nm, and one for deuterium, 656.27 nm. We need to find the minimum number of slits required to resolve these in second order and calculate the angles and angular separation using a grating with 500 slits/mm.
02

Calculate the Required Slit Number

We use the formula for the resolving power \( R = \frac{m \cdot N}{\Delta \lambda/\lambda} \), where \( m \) is the order of the spectrum (2 in this case), \( N \) is the number of slits, \( \lambda \) is the average wavelength, and \( \Delta \lambda \) is the difference between the two wavelengths.Let \( \lambda_1 = 656.45 \text{ nm} \) and \( \lambda_2 = 656.27 \text{ nm} \). Then \( \Delta \lambda = \lambda_1 - \lambda_2 = 0.18 \text{ nm} \).The average wavelength \( \lambda = \frac{\lambda_1 + \lambda_2}{2} = \frac{656.45 \text{ nm} + 656.27 \text{ nm}}{2} = 656.36 \text{ nm} \).Calculate: \( R = \frac{2 \cdot N}{0.18/656.36} \).Solve for \( N \): \( N = \frac{0.18 \cdot 656.36}{2} \approx 328.18 \). Since \( N \) must be a whole number, we round up to 329.
03

Calculate the Angles using Grating Equation

Use the grating equation for angles: \( d \cdot \sin \theta = m \cdot \lambda \), where \( d \) is the distance between two slits and \( m = 2 \).First, find \( d \), the distance between the slits: \( d = \frac{1}{500 \times 10^3} \text{ mm} = 2 \times 10^{-6} \text{ mm} \).For hydrogen (\( \lambda = 656.45 \text{ nm} = 656.45 \times 10^{-9} \text{ m} \)), solve for \( \theta_1 \):\[\sin \theta_1 = \frac{2 \times 656.45 \times 10^{-9}}{2 \times 10^{-6}} \approx 0.65645.\]\( \theta_1 = \arcsin(0.65645) \approx 41.15^\circ \).For deuterium (\( \lambda = 656.27 \text{ nm} = 656.27 \times 10^{-9} \text{ m} \)), solve for \( \theta_2 \):\[\sin \theta_2 = \frac{2 \times 656.27 \times 10^{-9}}{2 \times 10^{-6}} \approx 0.65627.\]\( \theta_2 = \arcsin(0.65627) \approx 41.13^\circ \).
04

Calculate the Angular Separation

The angular separation \( \Delta \theta \) is \( \theta_1 - \theta_2 \).\( \Delta \theta \approx 41.15^\circ - 41.13^\circ = 0.02^\circ \).

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Key Concepts

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

Resolving Power
Resolving power is a crucial concept in the field of spectroscopy. It determines the ability of an optical instrument, like a diffraction grating, to separate two close wavelengths. The mathematical expression for resolving power is given by the formula:
  • \( R = \frac{m \cdot N}{\Delta \lambda / \lambda} \)
Here, \( m \) represents the order of the spectrum, \( N \) is the number of slits in the grating, and \( \Delta \lambda \) is the wavelength difference that needs to be resolved compared to an average wavelength \( \lambda \). A high resolving power means the instrument can distinguish between very closely spaced wavelengths. This is essential when observing spectral lines of different isotopes in spectroscopy.

In our problem, resolving these two close wavelengths from hydrogen and deuterium requires determining the minimum slit number necessary to achieve the needed resolving power. By calculating, we find that at least 329 slits are required for the grating at the second order of the spectrum. This ensures that the subtle difference between the two isotopic lines is discernible.
Wavelength Resolution
Wavelength resolution refers to the smallest difference in wavelengths that can be detected by an optical instrument. It is a critical parameter, especially in spectroscopy, where distinguishing between small differences is often needed. For instance, isotopes of the same element will emit light at slightly different wavelengths, and achieving a high wavelength resolution is key to identifying these differences.

In solving our problem, the wavelength resolution needed is determined by the difference in the given wavelengths for hydrogen and deuterium, which is \( 0.18 \text{ nm} \). This value can be quantified in terms of resolving power as shown by the relation:
  • \( \Delta \lambda = \frac{\lambda_1 - \lambda_2}{2} \)
Achieving this resolution allows us to use the grating equations to compute the necessary angular distinctions, ensuring that small differences in wavelengths are evident in the spectrum as different angles of diffraction.
Spectroscopy
Spectroscopy is the study of how different wavelengths of light are absorbed, emitted, or scattered by materials. In the context of our exercise, spectroscopy helps us understand the differences in light emission between the isotopes of hydrogen, namely protium and deuterium.

The diffraction grating plays a pivotal role here, separating the close spectral lines of the isotopes by diffracting light through numerous slits. Each slit allows the light waves to interfere with each other, resulting in a pattern that is based on wavelength. By resolving such close spectral lines, we gain insights into the atomic structures and compositions of elements.

The ability of a diffraction grating to resolve such close lines is central to many practical applications in chemistry and physics, including the analysis of stars' spectra, determining elemental concentrations, and identifying materials based on their unique spectral "fingerprints."
Optics
Optics is the branch of physics dealing with light and its interactions. In our context, it involves understanding how light behaves as it encounters objects like diffraction gratings. Optics allows us to predict how light will diffract and the resulting interference patterns that form.

Using the grating equation:
  • \( d \cdot \sin \theta = m \cdot \lambda \)
we can calculate the angles at which different wavelengths will diffract. Here, \( d \) represents the distance between slits, which inversely affects the angle. The determination of angles \( \theta \) for the wavelengths of hydrogen and deuterium is crucial in resolving their spectra.

This part of optics showcases how precise manipulation of light based on the principles of diffraction and interference can lead to practical solutions in resolving closely spaced lines in a spectrum, which is fundamental for identifying minute differences in light properties.

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Most popular questions from this chapter

Visible light passes through a diffraction grating that has 900 slits/cm, and the interference pattern is observed on a screen that is 2.50 m from the grating. (a) Is the angular position of the first-order spectrum small enough for sin \(\theta \approx \theta\) to be a good approximation? (b) In the first- order spectrum, the maxima for two different wavelengths are separated on the screen by 3.00 mm. What is the difference in these wavelengths?

A slit 0.360 mm wide is illuminated by parallel rays of light that have a wavelength of 540 nm. The diffraction pattern is observed on a screen that is 1.20 m from the slit. The intensity at the center of the central maximum \((\theta = 0^\circ)\) 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?

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

The wavelength range of the visible spectrum is approximately 380-750 nm. White light falls at normal incidence on a diffraction grating that has 350 slits/mm. Find the angular width of the visible spectrum in (a) the first order and (b) the third order. (\(Note\): An advantage of working in higher orders is the greater angular spread and better resolution. A disadvantage is the overlapping of different orders, as shown in Example 36.4.)

Although we have discussed single-slit diffraction only for a slit, a similar result holds when light bends around a straight, thin object, such as a strand of hair. In that case, \(a\) is the width of the strand. From actual laboratory measurements on a human hair, it was found that when a beam of light of wavelength 632.8 nm was shone on a single strand of hair, and the diffracted light was viewed on a screen 1.25 m away, the first dark fringes on either side of the central bright spot were 5.22 cm apart. How thick was this strand of hair?

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