Select the correct answer.
Fred points a toy laser gun at a wall. Considering that the frequency of the gun's light is [tex]$4.91 \times 10^{14}$[/tex] hertz and that Planck's constant is [tex]$6.61 \times 10^{-34}$[/tex] joule seconds, what is the energy of the laser light?
A. [tex]$3.24 \times 10^{-19}$[/tex] joules
B. [tex]$4.91 \times 10^{15}$[/tex] joules
C. [tex]$2.04 \times 10^{-21}$[/tex] joules
D. [tex]$4.92 \times 10^{-18}$[/tex] joules
Answer (2)
Use the formula E = h f to calculate the energy of the laser light.
Substitute the given values: h = 6.61 × 1 0 − 34 Js and f = 4.91 × 1 0 14 Hz.
Calculate the energy: E = ( 6.61 × 1 0 − 34 ) × ( 4.91 × 1 0 14 ) = 3.24 × 1 0 − 19 J.
The energy of the laser light is 3.24 × 1 0 − 19 joules .
Explanation
Understanding the Problem We are given the frequency of the laser light and Planck's constant, and we need to find the energy of the laser light. We can use the formula E = h f , where E is the energy, h is Planck's constant, and f is the frequency.
Given Values We are given: Frequency, f = 4.91 × 1 0 14 Hz Planck's constant, h = 6.61 × 1 0 − 34 Js
Applying the Formula Now, we substitute the given values into the formula E = h f :
E = ( 6.61 × 1 0 − 34 Js ) × ( 4.91 × 1 0 14 Hz )
Calculating the Energy Calculating the energy: E = 6.61 × 4.91 × 1 0 − 34 + 14 J E = 32.4551 × 1 0 − 20 J E = 3.24551 × 1 0 − 19 J Rounding to two decimal places, we get: E = 3.24 × 1 0 − 19 J
Selecting the Correct Answer Comparing our result with the given options, we see that the correct answer is A. 3.24 × 1 0 − 19 joules.
Examples
Lasers are used in many applications, from barcode scanners to laser pointers. Understanding the energy of laser light is crucial in designing and using these devices safely and effectively. For example, in laser eye surgery, the energy of the laser must be precisely controlled to avoid damaging the eye. Similarly, in laser cutting, the energy must be high enough to cut through the material, but not so high that it damages the surrounding area. This calculation demonstrates a fundamental concept in quantum mechanics and has practical applications in various fields.
The energy of the laser light emitted from Fred's toy laser gun is calculated using the formula E = h f . By substituting Planck's constant and the frequency of the light, we find that the energy equals 3.24 × 1 0 − 19 joules, which corresponds to option A.
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