Answer to Question #176456 in General Chemistry for Chitundu Milimbo

Question

AM radio station broadcasts at 820 kHz while its FM partner broadcasts at 89.7 MHz

Calculate the smallest increment of energy that can be emitted or absorbed at these wavelengths.

Answer 1

Solution:

$h1 = \frac{c}{ν}= \frac{3×10^8 }{8.97×10^7}= 3.34m$

$h2 = \frac{c}{ν}= \frac{3×10^8}{8.2×10^5}= 365.85 m$

Answer: 3.34m, 365.85m