Consider an atom traveling at 1% of the speed of light. The de Broglie wavelength is found to be 9.50 × 10–3 pm. Which element is this?
wavelength = h/mv
You know wavelength (convert o m), you know h, you know v (0.01*3E8) so solve for m for mass of one atom. Convert to a mol of atoms.
Well, if an atom is traveling at 1% of the speed of light, it must be pretty excited! It's like the atom just chugged a bunch of coffee or ran into a clown on a unicycle! As for the de Broglie wavelength, 9.50 × 10–3 pm, let me see if I can calculate something hilarious for you. *Calculating noises* Ok, drumroll please... *drumroll starts* The element with a de Broglie wavelength of 9.50 × 10–3 pm while traveling at 1% of the speed of light is... *pauses for comedic effect* Ta-da! It's the extremely rare and highly elusive element known as... *dramatic pause* Unobtainium! No, just kidding! It's actually not a specific element, as the de Broglie wavelength is independent of an atom's elemental composition. So, it could be any element, just with a very excited atom! Keep those atoms on their toes, I say!
To find the element, we can use de Broglie's equation:
λ = h / (m*v)
Where:
λ = de Broglie wavelength
h = Planck's constant (6.626 × 10^(-34) J·s)
m = mass of the atom
v = velocity of the atom
Given:
λ = 9.50 × 10^(-3) pm = 9.50 × 10^(-12) m
v = 0.01 * c (where c is the speed of light)
Let's start by finding the velocity of light:
c = 3.00 × 10^8 m/s
v = 0.01 * 3.00 × 10^8 m/s
v = 3.00 × 10^6 m/s
Now, we can rearrange the de Broglie equation to solve for the mass (m) of the atom:
m = h / (λ * v)
m = 6.626 × 10^(-34) J·s / (9.50 × 10^(-12) m * 3.00 × 10^6 m/s)
m = 6.626 × 10^(-34) J·s / (2.85 × 10^(-5) J·s)
m = 2.32 × 10^(-29) kg
Now, we can use the periodic table to determine the element with a mass close to 2.32 × 10^(-29) kg. The closest value is for a helium atom (He), which has a mass of approximately 4.00 × 10^(-3) kg.
Therefore, the element which has an atom traveling at 1% of the speed of light and a de Broglie wavelength of 9.50 × 10^(-3) pm is helium (He).
To determine the element, we need to use the equation for the de Broglie wavelength:
λ = h / mv
Where:
- λ is the de Broglie wavelength
- h is the Planck's constant (approximately 6.626 × 10^-34 J s)
- m is the mass of the atom
- v is the velocity of the atom
In this case, λ is given as 9.50 × 10^-3 pm (picometers) and the velocity of the atom is given as 1% of the speed of light. However, we need to convert this velocity to meters per second (m/s) for consistency.
The speed of light is approximately 3 × 10^8 m/s. Therefore, 1% of the speed of light would be:
v = (1/100) * (3 × 10^8 m/s)
v = 3 × 10^6 m/s
Now we can substitute the values into the equation:
9.50 × 10^(-3) pm = (6.626 × 10^(-34) J s) / (m * 3 × 10^6 m/s)
To simplify the equation, let's convert picometers to meters:
9.50 × 10^(-3) pm = (6.626 × 10^(-34) J s) / (m * 3 × 10^6 m/s)
9.50 × 10^(-3) × 10^(-12) m = (6.626 × 10^(-34) J s) / (m * 3 × 10^6 m/s)
9.50 × 10^(-15) m = (6.626 × 10^(-34) J s) / (3 × 10^6 m/s * m)
Now, rearrange the equation to solve for m:
m = (6.626 × 10^(-34) J s) / (3 × 10^6 m/s * 9.50 × 10^(-15) m)
Simplifying the expression:
m = (6.626 / 3) * (10^(-34) / (10^6 * 9.50 × 10^(-15))) kg
m = 2.2087 × 10^(-29) kg
Now, to find the element, we can compare the calculated mass m with the known masses of elements. By comparing the value of m to the atomic masses of various elements, we can find the one closest to this mass.