Answer :
Electrons must be accelerated through Part A: a potential difference of 66.9 V to have the same wavelength as an x-ray of wavelength 0.150 nm. Part B: Electrons must be accelerated through a potential difference of 41.8 V.
Part A: The de Broglie wavelength of a particle can be calculated using the equation:
λ = h / √(2 * m * e * ΔV),
where λ is the wavelength, h is Planck's constant, m is the mass of an electron, e is the charge on an electron, and ΔV is the potential difference.
We can rearrange this equation to solve for ΔV:
ΔV = (h / λ)² / (2 * m * e).
Substituting the given values, we get:
ΔV = (6.63×10⁻³⁴ J⋅s / (0.150×10⁻⁹ m))² / (2 * 9.11×10⁻³¹ kg * 1.60×10⁻¹⁹ C).
Evaluating this expression, we find that ΔV is approximately 66.9 V.
Part B: The kinetic energy of an electron can be calculated using the equation:
E = (1/2) * m * v²,
where E is the energy, m is the mass of an electron, and v is the velocity.
The velocity of an electron can be found using the equation:
v = √(2 * e * ΔV / m),
where ΔV is the potential difference.
Substituting the given values, we have:
v = √(2 * 1.60×10⁻¹⁹ C * ΔV / 9.11×10⁻³¹ kg).
To find the potential difference ΔV that gives the same energy as the x-ray in Part A, we need to equate the kinetic energy of the electron to the energy of the x-ray, which is given by E = hc / λ, where c is the speed of light in a vacuum.
Setting these two expressions equal to each other and solving for ΔV, we find:
ΔV = (h * c)² / (2 * e * λ).
Substituting the given values, we get:
ΔV = (6.63×10⁻³⁴ J⋅s * 3.00×10⁸m/s)² / (2 * 1.60×10⁻¹⁹ C * 0.150×10⁻⁹ m).
Calculating this expression, we find that ΔV is approximately 41.8 V.
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