Final answer:
The kinetic energy of a bare helium nucleus at 2.00% of the speed of light is 5.312 × 10–¹² J. This is equivalent to 3.320 × 10⁷ eV. To obtain this energy, a voltage of 1.327 × 10^–¹² V is needed.
Step-by-step explanation:
To calculate the kinetic energy of a bare helium nucleus at 2.00% of the speed of light, we can use the non-relativistic formula for kinetic energy: KE = (1/2)mv^2.
Given that the mass of the helium nucleus is 6.64 × 10–²⁷ kg and the velocity is 2.00% of the speed of light, we can calculate the kinetic energy:
KE = (1/2)(6.64 × 10–²⁷ kg)(2.00%c)^2
where c is the speed of light (3.00 × 10^8 m/s).
Substituting the values, we get:
KE = (1/2)(6.64 × 10–²⁷ kg)(0.02c)^2 = 5.312 × 10–¹² J
To convert this energy to electron volts (eV), we can use the conversion factor: 1 eV = 1.602 × 10–¹⁹ J. Therefore, the kinetic energy in electron volts is:
KE (eV) = (5.312 × 10–¹² J)/(1.602 × 10–¹⁹ J/eV) = 3.320 × 10⁷ eV
To calculate the voltage needed to obtain this energy, we can use the equation: V = KE/q, where V is the voltage, KE is the kinetic energy, and q is the charge of the helium nucleus (2 times the elementary charge: 2e).
Substituting the values, we get:
V = (5.312 × 10–¹² J)/(2e) = 1.327 × 10^–¹² V