An e−1 is released by an electron microscope at 3.00 × 106 m/s.
Determine in meters using the de Broglie relation
Substitute kgˑm2/s2 for J and show all units.
Then, convert to pm and show the conversion factor.
Determine the kinetic energy using EK = (1/2)mv2, where v is velocity.
Show all units.
So would it be 9.10938 x10^-31 j Gould would you convert that?
I'm a little unsure of what you want. I assume that "box" I see and can't read is lamda. lambda = h/mv
So you plug in h, you have m right for the electron, and you have v listed, solve for wavelength.
If you want KE use KE = 1/2 mv^2 and units will be in J.
If you want to convert lambda in meters to pm, that is ?m x (10^12 pm/m) = ?
To determine the value of λ (wavelength) using the de Broglie relation, we can use the equation:
λ = h / p
Where:
λ = wavelength
h = Planck's constant (6.62607015 × 10^-34 J·s)
p = momentum of the particle
To find the momentum of the particle, we can use the equation:
p = mv
Where:
m = mass of the particle (in kg)
v = velocity of the particle (in m/s)
Given that the velocity of the e^-1 is 3.00 × 10^6 m/s, we need to find the mass of an electron.
The mass of an electron is approximately 9.10938356 × 10^-31 kg.
Now we can calculate the momentum:
p = (9.10938356 × 10^-31 kg) × (3.00 × 10^6 m/s) = 2.732815068 × 10^-24 kg·m/s
Finally, substituting the values of h and p into the wavelength equation:
λ = (6.62607015 × 10^-34 J·s) / (2.732815068 × 10^-24 kg·m/s)
To express λ in meters, we can do the unit conversion:
λ = 6.62607015 × 10^-34 J·s / 2.732815068 × 10^-24 kg·m/s
= 2.42440035 × 10^-10 m
To convert λ to picometers (pm), we can use the conversion factor:
1 m = 10^12 pm
Converting λ to picometers:
λ = 2.42440035 × 10^-10 m × (10^12 pm / 1 m)
= 2.42440035 × 10^2 pm
Therefore, the wavelength λ is approximately 242.44 pm.
Now, to determine the kinetic energy (EK) using the equation EK = (1/2)mv^2:
EK = (1/2) × (9.10938356 × 10^-31 kg) × (3.00 × 10^6 m/s)^2
Simplifying:
EK = (1/2) × 9.10938356 × 10^-31 kg × (3.00 × 10^6 m/s)^2
= 4.103222602 × 10^-18 J
So, the kinetic energy EK is approximately 4.103222602 × 10^-18 J.