Final answer:
The wavelength of the proton is approximately 0.826 picometers (pm). The kinetic energy of the proton is approximately 2.94 mega-electron volts (MeV). The equivalent voltage through which the proton was accelerated is approximately 1.23 million volts (MV).
Step-by-step explanation:
To calculate the wavelength of a proton traveling at 25.0% of the speed of light, we can use the de Broglie wavelength equation: λ = h / (mv), where λ is the wavelength, h is Planck's constant (6.626 × 10⁻³⁴ J s), m is the mass of the proton (1.673 × 10⁻²⁷ kg), and v is the velocity of the proton. Plugging in the values, we get: λ = (6.626 × 10⁻³⁴ J s) / ((1.673 × 10⁻²⁷ kg)(0.25c)), where c is the speed of light. Solving this equation gives us a wavelength of approximately 0.826 picometers (pm).
To calculate the kinetic energy of the proton assuming it is nonrelativistic, we can use the equation KE = (1/2)mv², where KE is the kinetic energy, m is the mass of the proton, and v is the velocity of the proton. Plugging in the values, we get: KE = (1/2)(1.673 × 10⁻²⁷kg)(0.25c)². Solving this equation gives us a kinetic energy of approximately 2.94 mega-electron volts (MeV).
To calculate the equivalent voltage through which the proton was accelerated, we can use the equation PE = qV, where PE is the potential energy, q is the charge of the proton, and V is the voltage. Since the proton is accelerated to 25.0% of the speed of light, we can assume that it gained its kinetic energy in only one pass through the potential difference. Therefore, the potential energy is equal to the kinetic energy. Plugging in the values, we get: (1.602 × 10⁻¹⁹ C)V = (1/2)(1.673 × 10⁻²⁷ kg)(0.25c)² Solving this equation gives us a voltage of approximately 1.23 million volts (MV).