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Chapter 16: Problem 30

Express the wavelength 2500 脜 in micrometers and nanometers.

Short Answer

Expert verified
2500 脜 = 0.25 渭m and 250 nm.

Step by step solution

01

Understand the Units

Before solving the problem, recognize the units involved. 脜ngstr枚m (脜) is a unit of length equal to \(10^{-10}\) meters. We are converting this to micrometers (\(\mu m\)), where 1 micrometer equals \(10^{-6}\) meters, and to nanometers (nm), where 1 nanometer equals \(10^{-9}\) meters.
02

Convert 脜ngstr枚ms to Micrometers

To convert 2500 脜 to micrometers, first convert 脜ngstr枚ms to meters: \[ 2500 \, 脜 = 2500 \, \times \, 10^{-10} \, \text{meters} = 2.5 \, \times \, 10^{-7} \, \text{meters} \]Then, convert meters to micrometers by multiplying by \(10^6\): \[ 2.5 \, \times \, 10^{-7} \, \text{meters} = 2.5 \, \times \, 10^{-7} \, \times \, 10^6 \, \text{micrometers} = 0.25 \, \mu m \]
03

Convert 脜ngstr枚ms to Nanometers

To convert 2500 脜 to nanometers, first convert 脜ngstr枚ms to meters:\[ 2500 \, 脜 = 2500 \, \times \, 10^{-10} \, \text{meters} = 2.5 \, \times \, 10^{-7} \, \text{meters} \]Then, convert meters to nanometers by multiplying by \(10^9\): \[ 2.5 \, \times \, 10^{-7} \, \text{meters} = 2.5 \, \times \, 10^{-7} \, \times \, 10^9 \, \text{nanometers} = 250 \, \text{nm} \]

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Key Concepts

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

脜ngstr枚m Conversion
The 脜ngstr枚m is a convenient unit often used in fields like physics and chemistry to express very small lengths, particularly in the atomic scale. To fully understand and apply 脜ngstr枚m conversion, it's essential to recognize that 1 脜ngstr枚m (脜) is defined as 10鈦宦光伆 meters. This tiny unit is incredibly helpful when working with wavelengths of light or distances between atoms in a molecule.

When converting 脜ngstr枚ms into other units, always start by expressing the value in meters. In the exercise example, 2500 脜 is first converted to meters:
  • 2500 脜 = 2500 脳 10鈦宦光伆 meters = 2.5 脳 10鈦烩伔 meters
By understanding and memorizing the basic length conversions (like 1 脜 = 10鈦宦光伆 meters), you can effortlessly switch between different units used in scientific measurements.
Micrometer Conversion
Often seen in biological sciences and engineering, the micrometer ( 渭m) is another small unit of length equating to 10鈦烩伓 meters. Converting 脜ngstr枚ms to micrometers involves an intermediate step of first converting 脜ngstr枚ms to meters, which was calculated previously as 2.5 脳 10鈦烩伔 meters.

To convert meters to micrometers, multiply by 10鈦� because:
  • 1 micrometer = 10鈦烩伓 meters
So, you multiply the 2.5 脳 10鈦烩伔 meters by 10鈦�:
  • 2.5 脳 10鈦烩伔 meters 脳 10鈦� = 0.25 渭m
This conversion highlights the need to adjust the power of ten. Understanding these powers is crucial when working with any metric conversions, especially in contexts that require very precise measurements.
Nanometer Conversion
Nanometers (nm) are commonly used in the world of physics, chemistry, and technology, especially for small-scale measurements like those of wavelengths or tiny components in electronic devices. Usually abbreviated as nm, 1 nanometer equals 10鈦烩伖 meters.

When converting from 脜ngstr枚ms to nanometers, first convert the 脜ngstr枚ms to meters鈥攋ust as in the previous sections:
  • 2500 脜 = 2.5 脳 10鈦烩伔 meters
Then, with the known conversion factor:
  • 1 nanometer = 10鈦烩伖 meters
Multiply the value in meters by 10鈦� to get nanometers:
  • 2.5 脳 10鈦烩伔 meters 脳 10鈦� = 250 nm
By practicing these kinds of conversions, the process becomes intuitive. The key is recognizing how to handle the powers of ten, ensuring accurate and smooth conversions between these very small units.

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Most popular questions from this chapter

Why do acid-base base indicators change color in going from acid to alkaline solution?

Define absorptivity and molar absorptivity.

Distinguish between the two types of monochromators (light dispersers) used in spectrophotometers and list the advantages and disadvantages of each.

Iron(II) is determined spectrophotometrically by reacting with 1,10 -phenanthroline to produce a complex that absorbs strongly at \(510 \mathrm{nm}\). A stock standard iron(II) solution is prepared by dissolving 0.0702 g ferrous ammonium sulfate, \(\mathrm{Fe}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}\), in water in a 1-L volumetric flask, adding \(2.5 \mathrm{~mL}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}\), and diluting to volume. A series of working standards is prepared by transferring \(1.00-, 2.00-, 5.00-,\) and \(10.00-\mathrm{mL}\) aliquots of the stock solution to separate \(100-\mathrm{mL}\) volumetric flasks and adding hydroxylammonium chloride solution to reduce any iron(III) to iron(II), followed by phenanthroline solution and then dilution to volume with water. A sample is added to a 100 -mL volumetric flask and treated in the same way. A blank is prepared by adding the same amount of reagents to a 100 -mL volumetric flask and diluting to volume. If the following absorbance readings measured against the blank are obtained at \(510 \mathrm{nm}\), how many milligrams iron are in the sample? (Table)

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