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4. Calculate the wavelength of electromagnetic radiation if the frequency is 2.50 x 1014 Hz?

- anonymous

- schrodinger

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- anonymous

|dw:1443734331234:dw|

- anonymous

v = frequency
but i don't know what m is.

- anonymous

and h is planks 6.626 x 10^-34

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## More answers

- Photon336

@theonefatcat I would like to show you this problem step by step.
the speed of light is equivalent to the speed of light times the frequency.
\[c = \lambda *v \]
we ask ourselves what we are looking for? we need wavelength
so we rearrange our formula by isolating lambda like this
|dw:1443734324536:dw|
now after we do that we know that the speed of light c = 2.99x10^8 m/s
and frequency is in 1/s so the units should be this.
\[ \frac{ c }{ v } = \frac{ m/s }{ 1/s } = \frac{ m(s) }{ (s) } = m = \lambda \]

- Photon336

sorry speed of light = wavelength times frequency

- anonymous

ok so see where you're going but i don't really understand the units like m/s

- Photon336

@Theonefatcat the reason why I did it like that was to show you the units we need.
wavelength is in meters so the unit we need to find is meters that's to show you that this is the right equation to use. always remember to check the units too

- Photon336

m/s = meters per second. that's how we measure speed. speed of light is in meters per second. and hertz is 1/seconds or seconds^-1

- anonymous

|dw:1443734615005:dw|

- anonymous

|dw:1443734721890:dw|

- Photon336

yep :)
\[\frac{ c }{ v } = \frac{ 2.99x10^{8} (m/s) }{ 2.50x10^{14} (1/s) } = \lambda \]

- anonymous

|dw:1443734798401:dw|

- anonymous

my teacher just tells us to round speed of light to 3 btw

- Photon336

|dw:1443734899109:dw|

- Photon336

@Theonefatcat yeah it's better if you round it to 3 i just took the exact number.

- anonymous

ok so the answer is 1.2 x 10^-6 m

- Photon336

@theonefatcat yep makes sense to me
FYI
|dw:1443735021493:dw|

- anonymous

ok ty

- Photon336

Yep no problem! remember like i did checking your units is such an important part to solving these problems, always ask yourself what units are wavelength in, meters and then check to see if those units cancel out to give you what you need in the equation.

- anonymous

i will i have a big test tomorrow you really helped. He takes points of for not having the units.

- Photon336

When I first did this i thought it was stupid to do that but it's actually very important. if you need meters and you end up with kilograms something is wrong. so yeah

- anonymous

i might have another question when i get to a part where they ask about neutrons and photons.

- anonymous

so i'll ask again later.

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