Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 6 - Electronic Structure of Atoms - Exercises: 6.19d

Answer

$1.5\times10^4m$

Work Step by Step

*Strategy: We would use the classic formula of the relationship between distance, time and velocity: $v\times t=x$ in which $v$: velocity of object $t$: the time the object travels $x$: the distance the object covers during the time $t$ and with velocity $x$ - Step 1: Find the known quantities We know that electromagnetic radiation always travels with the speed of light in vacuum, so $v = c \approx 2.998\times10^8m/s$ We also know that the electromagnetic radiation travels in $50.0\mu s$, so $t = 50.0\mu s = 5\times10^{-5}s$ (remember to change the unit to second ($s$) to be compatible with $v$ for calculation) - Step 2: Do the calculation The distance the electromagnetic radiation travels is $x=v\times t= (2.998\times10^8)\times(5\times10^{-5})\approx1.5\times10^4m$
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