An experiment has four possible outcomes, labeled A to D. The probability of A is PA = 40% and of B is PB = 30%. Outcome C is twice as probable as outcome D. What are the probabilities PC and PD?
Ch 39: Wave Functions and Uncertainty
39장, 문제 8
In one experiment, 2000 photons are detected in a 0.10-mm-wide strip where the amplitude of the electromagnetic wave is 10 V/m. How many photons are detected in a nearby 0.10-mm-wide strip where the amplitude is 30 V/m?
검증된 단계별 안내1
Understand that the number of photons detected is proportional to the intensity of the electromagnetic wave, and the intensity is proportional to the square of the amplitude of the wave. This means \( I \propto A^2 \), where \( I \) is the intensity and \( A \) is the amplitude.
Write the relationship between the number of photons detected \( N \) and the amplitude \( A \): \( N \propto A^2 \). This implies \( \frac{N_2}{N_1} = \frac{A_2^2}{A_1^2} \), where \( N_1 \) and \( N_2 \) are the number of photons detected in the first and second strips, and \( A_1 \) and \( A_2 \) are their respective amplitudes.
Substitute the given values for \( A_1 \) and \( A_2 \): \( A_1 = 10 \, \text{V/m} \) and \( A_2 = 30 \, \text{V/m} \). The equation becomes \( \frac{N_2}{N_1} = \frac{(30)^2}{(10)^2} \).
Simplify the ratio: \( \frac{N_2}{N_1} = \frac{900}{100} = 9 \). This means the number of photons detected in the second strip is 9 times the number detected in the first strip.
Multiply the number of photons detected in the first strip by this ratio to find \( N_2 \): \( N_2 = 9 \times 2000 \). This gives the number of photons detected in the second strip.

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이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
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주요 개념
질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.
Photon Detection
Photon detection refers to the process of identifying and counting individual photons, which are the fundamental particles of light. The number of detected photons can be influenced by factors such as the intensity of the light source and the area over which the detection occurs. In this context, the experiment measures how varying the amplitude of the electromagnetic wave affects the number of photons detected in a specified area.
추천 영상:
가이드 코스
Example 2
Amplitude of Electromagnetic Waves
The amplitude of an electromagnetic wave is a measure of the maximum electric field strength of the wave. It is directly related to the intensity of the wave, which is proportional to the square of the amplitude. Higher amplitude indicates a stronger wave, which can lead to a greater number of photons being emitted or detected, as seen in the comparison between the two strips in the experiment.
추천 영상:
가이드 코스
Introduction to Electromagnetic (EM) Waves
Intensity and Photon Flux
Intensity in the context of electromagnetic waves is defined as the power per unit area and is proportional to the square of the amplitude. Photon flux, which is the number of photons passing through a unit area per unit time, increases with higher intensity. Therefore, by understanding the relationship between intensity, amplitude, and photon detection, one can calculate the expected number of photons detected in different conditions.
추천 영상:
가이드 코스
Wave Intensity
관련 실천
교과서 질문
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교과서 질문
FIGURE EX39.13 shows the probability density for an electron that has passed through an experimental apparatus. What is the probability that the electron will land in a 0.010-mm-wide strip at x = 0.000 mm?
1453
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교과서 질문
When 5×1012 photons pass through an experimental apparatus, 2.0×109 land in a 0.10-mm-wide strip. What is the probability density at this point?
98
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교과서 질문
FIGURE EX39.12 shows the probability density for an electron that has passed through an experimental apparatus. If 1.0×106 electrons are used, what is the expected number that will land in a 0.010-mm-wide strip at 2.000 mm?
103
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