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The \(D\) line in the spectrum of sodium is a doublet with wavelengths \(589.0\) and \(589.6 \mathrm{~nm}\). Calculate the minimum number of lines needed in a grating that will resolve this doublet in the third-order spectrum.

Short Answer

Expert verified
328 lines are needed to resolve the doublet.

Step by step solution

01

Understand the concept of resolving power

To resolve two closely spaced wavelengths using a diffraction grating, the resolving power \( R \) of the grating must be greater than or equal to the ratio of one of the wavelengths \( \lambda \) to the difference between the wavelengths \( \Delta \lambda \). This is expressed as \( R \geq \frac{\lambda}{\Delta \lambda} \).
02

Calculate the difference in wavelengths

The difference in wavelengths for the sodium doublet is \( \Delta \lambda = 589.6 \, \text{nm} - 589.0 \, \text{nm} = 0.6 \, \text{nm} \).
03

Compute the resolving power required

Using the formula for resolving power, \( R = \frac{\lambda}{\Delta \lambda} \), and substituting \( \lambda = 589.0 \, \text{nm} \) and \( \Delta \lambda = 0.6 \, \text{nm} \), we calculate \( R = \frac{589.0}{0.6} \approx 981.67 \).
04

Relate resolving power to the grating equation

The resolving power \( R \) of a grating is also given by \( R = mN \), where \( m \) is the order of the spectrum (third order in this case) and \( N \) is the number of lines in the grating. We need to solve for \( N \).
05

Solve for the minimum number of lines

Given \( R = 981.67 \) and \( m = 3 \), use the formula \( R = mN \). Solving for \( N \), we get \( N = \frac{R}{m} = \frac{981.67}{3} \approx 327.22 \). Since \( N \) must be an integer, round up to the nearest whole number, giving \( N = 328 \).

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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 when it comes to diffraction grating. It measures the ability of the grating to separate two closely spaced wavelengths. Imagine you have two colors of light, very close in shade. To see them separately, the grating needs a certain resolving power. Resolving power (\( R \)) is calculated using the formula:\[ R = \frac{\lambda}{\Delta \lambda} \]where:
  • \( \lambda \) is the wavelength of light.
  • \( \Delta \lambda \) is the difference between the two wavelengths you're trying to resolve.
The larger the resolving power, the more capable the grating is at distinguishing between wavelengths that are close together. In our problem, resolving power ensures that the sodium doublet's closely spaced wavelengths can be differentiated from one another.
Spectrum Order
In the context of diffraction gratings, spectrum order refers to the different levels of diffraction that can occur when light passes through the grating. Each order corresponds to a different angle at which the diffracted light can be observed. For example, the first-order spectrum (\( m = 1 \)) is observed at the smallest angle, while higher orders like third-order (\( m = 3 \)) appear at larger angles. The order of the spectrum has a direct impact on the resolving power needed for separating two wavelengths. In our problem, we're dealing with the third-order spectrum, meaning we observe light that is diffracted in such a way that it has traveled three "loops" of the wavelength distance compared to just one in the first-order.
Wavelength Difference
The wavelength difference, denoted as \( \Delta \lambda \), is the gap between the two wavelengths you're aiming to resolve. It's a critical factor in determining the resolving power. The smaller this difference, the higher the resolving power required to separate them.In the sodium doublet example, the wavelengths are 589.0 nm and 589.6 nm. The difference is calculated as:\[ \Delta \lambda = 589.6 \, \text{nm} - 589.0 \, \text{nm} = 0.6 \, \text{nm} \]Understanding the wavelength difference helps in understanding why a high resolving power is necessary. It's all about pinpointing how much the wavelengths differ and how fine-tuned the equipment needs to be to distinguish between them.
Sodium Doublet
The sodium doublet is a famous example in the study of spectroscopy and diffraction. It consists of two closely spaced spectral lines, often found in sodium emission spectra, at wavelengths of 589.0 nm and 589.6 nm . These lines are almost indistinguishable without a proper resolving power. They're named "doublet" because they appear as two lines due to two different electronic transitions occurring in sodium atoms. Resolving these lines is crucial in spectroscopic analysis as it provides insight into the atomic structure and behavior of sodium. By using the correct number of lines in a diffraction grating, as calculated in our problem, these two lines can be separated clearly even though they lie close to one another.

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Most popular questions from this chapter

Nuclear-pumped \(x\)-ray lasers are seen as a possible weapon to destroy ICBM booster rockets at ranges up to \(1500 \mathrm{~km}\). One limitation on such a device is the spreading of the beam due to diffraction, with resulting dilution of beam intensity. Consider such a laser operating at a wavelength of \(1.40 \mathrm{~nm}\). The element that emits light is the end of a wire with diameter \(0.400 \mathrm{~mm}\). (a) Calculate the diameter of the central beam at a target \(2000 \mathrm{~km}\) away from the beam source. (b) What is the ratio of the beam intensity at the target to that at the end of the wire? (The laser is fired from space, so neglect any atmospheric absorption.)

Monochromatic light with wavelength \(420 \mathrm{~nm}\) is incident on a slit with width \(0.050 \mathrm{~mm}\). The distance from the slit to a screen is \(3.5 \mathrm{~m}\). Consider a point on the screen \(2.2 \mathrm{~cm}\) from the central maximum. Calculate (a) \(\theta\) for that point, (b) \(\alpha\), and (c) the ratio of the intensity at that point to the intensity at the central maximum.

A grating has 600 rulings/mm and is \(6.0 \mathrm{~mm}\) wide. (a) What is the smallest wavelength interval it can resolve in the third order at \(\lambda=500 \mathrm{~nm}\) ? (b) How many higher orders of maxima can be seen?

a) How many bright fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern if \(\lambda=550 \mathrm{~nm}, d=0.180 \mathrm{~mm}\), and \(a=30.0 \mu \mathrm{m}\) ? (b) What is the ratio of the intensity of the third bright fringe to the intensity of the central fringe?

Light of wavelength \(420 \mathrm{~nm}\) is incident normally on a diffraction grating. Two adjacent maxima occur at angles given by \(\sin \theta=0.2\) and \(\sin \theta=0.3\). The fourth-order maxima are missing. (a) What is the separation between adjacent slits? (b) What is the smallest slit width this grating can have? For that slit width, what are the (c) largest, (d) second largest, and (e) third largest values of the order number \(m\) of the maxima produced by the grating?

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