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\(\bullet\) Ultraviolet radiation. There are two categories of ultraviolet light. Ultraviolet A (UVA) has a wavelength ranging from 320 \(\mathrm{nm}\) to 400 nm. It is not so harmful to the skin and is necessary for the production of vitamin D. UVB, with a wavelength between 280 \(\mathrm{nm}\) and \(320 \mathrm{nm},\) is much more dangerous, because it causes skin cancer. (a) Find the frequency ranges of UVA and UVB. (b) What are the ranges of the wave numbers for UVA and UVB?

Short Answer

Expert verified
UVA frequency range: \(7.5 - 9.375\times10^{14}\) Hz; UVB: \(9.375 - 1.071\times10^{15}\) Hz. UVA wave numbers: \(2.5 - 3.125\times10^{6}\,\mathrm{m}^{-1}\); UVB: \(3.125 - 3.571\times10^{6}\,\mathrm{m}^{-1}\).

Step by step solution

01

Understanding the Relationship between Wavelength and Frequency

The frequency \( f \) of a wave is related to its wavelength \( \lambda \) by the formula \( c = f \lambda \), where \( c \) is the speed of light, approximately \( 3 \times 10^8 \, \mathrm{m/s} \). For this problem, we need to convert wavelengths from nanometers to meters by noting that \( 1 \, \mathrm{nm} = 10^{-9} \, \mathrm{m} \).
02

Calculating Frequency for UVA

For UVA, the wavelength ranges from \( 320 \, \mathrm{nm} = 320 \times 10^{-9} \, \mathrm{m} \) to \( 400 \, \mathrm{nm} = 400 \times 10^{-9} \, \mathrm{m} \). Using the formula \( f = \frac{c}{\lambda} \), the frequency range is calculated: - Lower Frequency: \( f_{\text{min}} = \frac{3 \times 10^8}{400 \times 10^{-9}} \approx 7.5 \times 10^{14} \, \mathrm{Hz} \)- Upper Frequency: \( f_{\text{max}} = \frac{3 \times 10^8}{320 \times 10^{-9}} \approx 9.375 \times 10^{14} \, \mathrm{Hz} \)
03

Calculating Frequency for UVB

For UVB, the wavelength ranges from \( 280 \, \mathrm{nm} = 280 \times 10^{-9} \, \mathrm{m} \) to \( 320 \, \mathrm{nm} = 320 \times 10^{-9} \, \mathrm{m} \). Calculate the frequency range:- Lower Frequency: \( f_{\text{min}} = \frac{3 \times 10^8}{320 \times 10^{-9}} \approx 9.375 \times 10^{14} \, \mathrm{Hz} \)- Upper Frequency: \( f_{\text{max}} = \frac{3 \times 10^8}{280 \times 10^{-9}} \approx 1.071 \times 10^{15} \, \mathrm{Hz} \)
04

Understanding Wave Numbers

The wave number \( u \) is the reciprocal of the wavelength and is given by \( u = \frac{1}{\lambda} \), usually expressed in the unit \( \mathrm{m}^{-1} \).
05

Calculating Wave Numbers for UVA

For UVA:- Lowest Wave Number: \( u_{\text{min}} = \frac{1}{400 \times 10^{-9}} \approx 2.5 \times 10^6 \, \mathrm{m}^{-1} \)- Highest Wave Number: \( u_{\text{max}} = \frac{1}{320 \times 10^{-9}} \approx 3.125 \times 10^6 \, \mathrm{m}^{-1} \)
06

Calculating Wave Numbers for UVB

For UVB:- Lowest Wave Number: \( u_{\text{min}} = \frac{1}{320 \times 10^{-9}} \approx 3.125 \times 10^6 \, \mathrm{m}^{-1} \)- Highest Wave Number: \( u_{\text{max}} = \frac{1}{280 \times 10^{-9}} \approx 3.571 \times 10^6 \, \mathrm{m}^{-1} \)

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Key Concepts

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

UVA and UVB
Ultraviolet (UV) radiation is a form of electromagnetic radiation that comes from the sun and artificial sources like tanning beds.
There are two major types of UV radiation: Ultraviolet A (UVA) and Ultraviolet B (UVB).

  • **UVA:** These rays have longer wavelengths, ranging from 320 nm to 400 nm. UVA penetrates deeper into the skin and is mainly associated with skin aging and wrinkles. However, it's not as harmful as UVB and plays a key role in the body's production of vitamin D.

  • **UVB:** With shorter wavelengths ranging from 280 nm to 320 nm, UVB rays are more energetic and harmful. They are primarily responsible for skin burns, skin cancer, and they can damage DNA in skin cells.
Understanding the effects and properties of these rays is crucial for protecting your skin, especially during prolonged sun exposure.
frequency calculation
The frequency of a wave is a critical property that determines its energy. For electromagnetic waves like UVA and UVB, the frequency is linked to the wavelength through the equation \[ c = f \lambda \]where:
  • \( c \) is the speed of light (approximately \( 3 \times 10^8 \) meters per second).

  • \( f \) represents the frequency.

  • \( \lambda \) is the wavelength.
For UVA, with wavelengths between \( 320 \) nm and \( 400 \) nm:- Convert these to meters by multiplying by \( 10^{-9} \).- The frequencies are calculated as:
\[ f_{min} = \frac{3 \times 10^8}{400 \times 10^{-9}} \approx 7.5 \times 10^{14} \, \text{Hz} \]\[ f_{max} = \frac{3 \times 10^8}{320 \times 10^{-9}} \approx 9.375 \times 10^{14} \, \text{Hz} \]For UVB, with wavelengths between \( 280 \) nm and \( 320 \) nm:- The frequencies range from:
\[ f_{min} = \frac{3 \times 10^8}{320 \times 10^{-9}} \approx 9.375 \times 10^{14} \, \text{Hz} \]\[ f_{max} = \frac{3 \times 10^8}{280 \times 10^{-9}} \approx 1.071 \times 10^{15} \, \text{Hz} \]Remember that higher frequency corresponds to higher energy, which makes UVB more harmful than UVA.
wave number
Wave number is another way to express the frequency of a wave, and it is defined as the number of wavelengths per unit distance, typically given in units of meters inverse (\( \text{m}^{-1} \)). This helps in analyzing the energy and absorption characteristics of electromagnetic waves.

The formula for wave number \( u \) is \[ u = \frac{1}{\lambda} \] where \( \lambda \) is the wavelength.For UVA, the wave numbers can be calculated by:- Minimum wave number:\[ u_{min} = \frac{1}{400 \times 10^{-9}} \approx 2.5 \times 10^6 \, \text{m}^{-1} \]- Maximum wave number:\[ u_{max} = \frac{1}{320 \times 10^{-9}} \approx 3.125 \times 10^6 \, \text{m}^{-1} \]For UVB:- Minimum wave number:\[ u_{min} = \frac{1}{320 \times 10^{-9}} \approx 3.125 \times 10^6 \, \text{m}^{-1} \]- Maximum wave number:\[ u_{max} = \frac{1}{280 \times 10^{-9}} \approx 3.571 \times 10^6 \, \text{m}^{-1} \]

This provides a convenient way to discuss similarities and differences between different types of UV radiation in terms of their potential impact.
wavelength conversion
When dealing with electromagnetic waves such as UVA and UVB rays, it becomes necessary to convert their wavelengths into different units. Most commonly, scientists need to convert the given wavelengths from nanometers (nm) to meters (m).
This conversion is crucial because it allows us to use standard equations involving the speed of light, such as the frequency and wave number calculations.

  • Since 1 nm is equivalent to \( 10^{-9} \) m, converting wavelengths from nanometers to meters is straightforward. For instance:

  • 320 nm becomes \( 320 \times 10^{-9} \) m and 280 nm becomes \( 280 \times 10^{-9} \) m.
This practice helps maintain consistency in measurements and ensures that calculations using formulas for frequency or wave number are accurate, as they rely on standard SI units. Understanding these conversions is essential for scientists and students involved in the study of light and radiation.

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Most popular questions from this chapter

\(\bullet\) The critical angle for total internal reflection at a liquid-air interface is \(42.5^{\circ} .\) (a) If a ray of light traveling in the liquid has an angle of incidence of \(35.0^{\circ}\) at the interface, what angle does the refracted ray in the air make with the normal? (b) If a ray of light traveling in air has an angle of incidence of \(35.0^{\circ}\) at the interface, what angle does the refracted ray in the liquid make with the normal?

\(\bullet\) The refractive index of a certain glass is \(1.66 .\) For what angle of incidence is light that is reflected from the surface of this glass completely polarized if the glass is immersed in (a) air or (b) water?

\(\bullet\) Light of original intensity \(I_{0}\) passes through two ideal polarizing filters having their polarizing axes oriented as shown in Figure \(23.62 .\) You want to adjust the angle \(\phi\) so that the intensity at point \(P\) is equal to \(I_{0} / 10 .\) (a) If the original light is unpolarized, what should \(\phi\) be? (b) If the original light is linearly polarized in the same direction as the polarizing axis of the first polarizer the light reaches, what should \(\phi\) be?

\(\bullet\) The intensity at a certain distance from a bright light source is 6.00 \(\mathrm{W} / \mathrm{m}^{2} .\) Find the radiation pressure (in pascals and in atmospheres) on (a) a totally absorbing surface and (b) a totally reflecting surface.

\(\bullet\) A glass plate having parallel faces and a refractive index of 1.58 lies at the bottom of a liquid of refractive index \(1.70 . \mathrm{A}\) ray of light in the liquid strikes the top of the glass at an angle of incidence of \(62.0^{\circ} .\) Compute the angle of refraction of this light in the glass.

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