В ядерной модели атома водорода электрон вращается вокруг ядра (протон) по круговой орбите, радиус которой R = 5,3 • 10-11 м. Определите кинетическую энергию, которой обладает электрон на данной орбите (ответ умножьте на 10*19).
The kinetic energy of the electron can be calculated using the formula:
KE = (mv^2)/2, where m is the mass of the electron and v is its velocity.
Since the electron is moving in a circular orbit, the velocity can be found using the formula for centripetal acceleration:
a = v^2/R, where R is the radius of the orbit.
Substituting the known values of R = 5.3 * 10^-11 m and the mass of an electron m = 9.11 * 10^-31 kg, we get:
v = √[(a * R^2)/m] = √[(9 * 10^9 * 1.602 * 10^-19 * 5.3 * 10^-11) / (5.11 * 10^-31)] = 2.19 *10^6 m/s
Thus, the kinetic energy of the electron is:
KE = (9.11 * 10^-31 * (2.19 * 10^6)^2)/2 = 2.43 * 10^-18 J
Multiplying this value by 10^19, as instructed in the prompt, we get the final answer of 2.43 * 10 J.
Therefore, the electron on the given orbit in the hydrogen atom has a kinetic energy of 2.43 * 10 J.
