Calculate the energies of the following waves in kilojoules per mole, and tell which member of each pair has the higher value. (a) An FM radio wave at 99.5 MHz and an AM radio wave at 1150 kHz

검증된 단계별 안내

Identify the formula to calculate the energy of a wave: \( E = h \nu \), where \( E \) is the energy, \( h \) is Planck's constant \( 6.626 \times 10^{-34} \text{ J s} \), and \( \nu \) is the frequency of the wave.

Convert the given frequencies from MHz and kHz to Hz. For FM radio wave: \( 99.5 \text{ MHz} = 99.5 \times 10^6 \text{ Hz} \). For AM radio wave: \( 1150 \text{ kHz} = 1150 \times 10^3 \text{ Hz} \).

Calculate the energy of one photon for each wave using the formula \( E = h \nu \). Substitute the frequency in Hz and Planck's constant into the equation for both FM and AM waves.

Convert the energy from joules per photon to kilojoules per mole. Use Avogadro's number \( 6.022 \times 10^{23} \text{ mol}^{-1} \) to find the energy per mole: \( E_{\text{mole}} = E_{\text{photon}} \times 6.022 \times 10^{23} \).

Compare the energies calculated for the FM and AM waves to determine which has the higher energy.

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이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.

영상 길이:

10m

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Electromagnetic Spectrum

The electromagnetic spectrum encompasses all types of electromagnetic radiation, which vary in wavelength and frequency. Radio waves, including FM and AM waves, are part of this spectrum and have longer wavelengths compared to visible light. Understanding the position of these waves within the spectrum is crucial for calculating their energy.

Energy of a Photon

The energy of a photon is directly related to its frequency and can be calculated using the equation E = hν, where E is energy, h is Planck's constant, and ν is frequency. This relationship indicates that higher frequency waves possess greater energy. Thus, comparing the energies of FM and AM waves requires converting their frequencies into energy values.

Conversion to Kilojoules per Mole

To express the energy of individual photons in kilojoules per mole, one must convert the energy calculated in joules to kilojoules and then multiply by Avogadro's number (approximately 6.022 x 10²³). This conversion is essential for comparing energies on a per-mole basis, which is a common practice in chemistry.