An electron is accelerated through an electric potential to a kinetic energy of 18.2 keV. What is its characteristic wavelength? (Hint: Recall that the kinetic energy of a moving object is E = 1/2 mv2 where m is the mass of the object and v is the speed of the object.)
To find the characteristic wavelength of an electron with a kinetic energy of 18.2 keV, we can use the de Broglie wavelength equation.
The de Broglie wavelength equation is given by:
λ = h / p
Where λ is the wavelength, h is Planck's constant (h = 6.626 x 10^-34 J.s), and p is the momentum of the electron.
First, we need to find the momentum of the electron. The momentum can be calculated using the formula:
p = √(2mE)
Where p is the momentum, m is the mass of the electron, and E is its kinetic energy.
The mass of an electron is approximately 9.10938356 × 10^-31 kg, and the kinetic energy is given as 18.2 keV. We need to convert the kinetic energy to joules.
1 keV = 1.602176634 × 10^-16 J
So, 18.2 keV = 18.2 × 1.602176634 × 10^-16 J = 2.918793733 × 10^-15 J.
Plugging these values into the formula:
p = √(2 × 9.10938356 × 10^-31 kg × 2.918793733 × 10^-15 J)
Calculating the square root:
p ≈ 5.278824767 × 10^-24 kg.m/s
Now that we have the momentum, we can calculate the wavelength:
λ = h / p
λ = (6.626 × 10^-34 J.s) / (5.278824767 × 10^-24 kg.m/s)
Calculating the division:
λ ≈ 1.255947826 × 10^-10 m
Therefore, the characteristic wavelength of the electron is approximately 1.255947826 × 10^-10 meters.
To find the characteristic wavelength of an electron, we can use the de Broglie wavelength equation, which relates the wavelength of a particle to its momentum:
λ = h / p
where λ is the wavelength, h is the Planck's constant (h = 6.63 x 10^-34 J·s), and p is the momentum of the electron.
The momentum of an object can be calculated using the equation:
p = m * v
where p is the momentum, m is the mass of the electron, and v is its velocity.
Given that the kinetic energy is 18.2 keV and the mass of an electron is approximately 9.11 x 10^-31 kg, we can first calculate the velocity of the electron.
Given:
E = 18.2 keV = 18.2 * 10^3 * 1.6 x 10^-19 J (converting keV to joules)
m = 9.11 x 10^-31 kg
v = ?
Using the kinetic energy equation:
E = 1/2 * m * v^2
Rearranging the equation to solve for v:
v^2 = 2 * E / m
v^2 = 2 * (18.2 * 10^3 * 1.6 x 10^-19) / (9.11 x 10^-31)
v^2 ≈ 5.96 x 10^12
Taking the square root of both sides:
v ≈ √(5.96 x 10^12)
v ≈ 2.44 x 10^6 m/s
Now, we can calculate the momentum:
p = m * v
p ≈ (9.11 x 10^-31) * (2.44 x 10^6)
p ≈ 2.23 x 10^-24 kg·m/s
Finally, we can calculate the wavelength using the de Broglie equation:
λ = h / p
λ ≈ (6.63 x 10^-34) / (2.23 x 10^-24)
λ ≈ 2.98 x 10^-10 m
So, the characteristic wavelength of the electron is approximately 2.98 x 10^-10 meters.