Answer:
Step-by-step explanation:
The de Broglie wavelength (λ) of a particle can be calculated using the following formula:
λ = h / p
Where:
λ = de Broglie wavelength
h = Planck's constant (6.626 x 10^-34 J·s)
p = momentum of the particle
In the given scenario, we have an electron with momentum p and energy E. We know that the energy of the electron can also be expressed in terms of its momentum and mass (m):
E = (p^2) / (2m)
Given that E = 1000 eV, we need to convert it to joules (J) because Planck's constant is in SI units (J·s) and the mass of the electron (m) is a constant.
1 eV = 1.602 x 10^-19 J
So, E = 1000 eV = 1000 * 1.602 x 10^-19 J = 1.602 x 10^-16 J
Now, we can find the momentum (p) using the energy (E) and mass (m) of the electron:
E = (p^2) / (2m)
1.602 x 10^-16 J = (p^2) / (2 * mass of electron)
Now, let's calculate p:
p^2 = 2 * 1.602 x 10^-16 J * mass of electron
p = sqrt(2 * 1.602 x 10^-16 J * mass of electron)
Now, we have the initial momentum (p) of the electron.
Next, you want to find the de Broglie wavelength (λ) using the initial momentum (p). You mentioned that λ = 0.04 nm in this scenario.
Now, if the speed of the electron increases by a factor of 10, the new momentum (p_new) will be 10 times the initial momentum (p):
p_new = 10 * p
Now, we can find the new de Broglie wavelength (λ_new) using the new momentum (p_new) and Planck's constant (h):
λ_new = h / p_new
Substitute the values:
λ_new = h / (10 * p)
λ_new = (1/10) * (h / p)
Since λ = 0.04 nm initially, and the speed is increased by a factor of 10:
λ_new = (1/10) * (0.04 nm) = 0.004 nm
So, the de Broglie wavelength of the electron will be 0.004 nm when its speed increases by a factor of 10.