I did a chem lab and now im not sure if im doing my calculations right .. Please help?

So i have to find out what ƒÉ will equal when i use the equation ƒÉ=d*(a/�ãa^2+b^2)

in my experiment i found out that a equals 108 cm, d and b were given to me .. so d=1.33*10^-6 m and b=2.0000 m

I plug the numbers in but they just don't seem right because im being asked for ă in cm then convert that number to nm find out frequence (s^-1) and energy (J/atom) but i think i might know how to do that .. i have eight more trials to go but this step is really holding be back , Please help me with just this exaple .. please help !

Of course, I'll be happy to help you with your calculations! Let's start by substituting the given values into the equation you provided:

ƒÉ = d * (a / �ã(a^2 + b^2))

a = 108 cm
d = 1.33 * 10^-6 m
b = 2.0000 m

First, it's important to convert the values to consistent units. Since you mentioned that you are asked for ă in cm, let's convert all the values to cm:

d = 1.33 * 10^-6 m = (1.33 * 10^-6 m) * (100 cm/1 m) = 1.33 * 10^-4 cm
b = 2.0000 m = (2.0000 m) * (100 cm/1 m) = 200.00 cm

Now we can substitute the values:

ƒÉ = (1.33 * 10^-4 cm) * (108 cm / �ã(108 cm^2 + 200.00 cm^2))

To simplify the denominator, we need to calculate the sum of the squares:

(108 cm^2 + 200.00 cm^2) = 11664 cm^2 + 40000 cm^2 = 51664 cm^2

Now we can substitute this value into the equation:

ƒÉ = (1.33 * 10^-4 cm) * (108 cm / �ã(51664 cm^2))

To simplify the denominator further, let's calculate the square root of 51664 cm^2:

�ã(51664 cm^2) ≈ 227.377 cm

Now we can substitute this value and simplify the expression:

ă = (1.33 * 10^-4 cm) * (108 cm / 227.377 cm)
≈ (1.33 * 10^-4 cm) * 0.4748
≈ 5.03152 * 10^-5 cm

So, the value of ă is approximately 5.03152 * 10^-5 cm based on the given values.

To convert this to nm (nanometers), we multiply it by 10 to the power of -7 (since 1 cm = 10^7 nm):

ă = (5.03152 * 10^-5 cm) * (10^-7 m/nm)
≈ 5.03152 * 10^-12 nm

Now we have ă in nm.

To find the frequency (s^-1) and energy (J/atom), we need additional information about the system or the element involved. Can you provide more details about the experiment or the element being studied?