1)A hypothetical neutral atom has 77 electrons. One of its electrons is held in circular orbit of radius r = 0.244 nm by electrostatic attraction between the electron and the nucleus. Assume: the interaction between electrons is negligible. Calculate the speed of orbit of this electron in m/s.
2)Assume the charge in the nucleus of an atom is evenly distributed through the nucleus.
In a hypothetical 56 electron neutral atom, the radius of the nucleus is 5.93 pm. Find the magnitude of the electric field at the surface of this nucleus, in N/C.
3)Two positive point charges are fixed on the x-axis: q1 is at the origin and q2 = 4.73q1 is at (14.6 cm, 0). You now place a third charge Q on the x-axis at (x,0) and it remains at rest. Find x in cm.
1) To calculate the speed of the electron in a circular orbit of radius r, we can use the formula for centripetal force. The electrostatic force between the electron and the nucleus provides the centripetal force.
The electrostatic force (F) between the electron and the nucleus is given by Coulomb's law:
F = (k * |e * q|) / r^2
where k is the electrostatic constant (approximated as 8.99 x 10^9 N*m^2/C^2), e is the elementary charge (1.6 x 10^-19 C), q is the charge of the nucleus (equal to the charge of the electron), and r is the radius of the orbit.
For a circular orbit, the centripetal force (F) is given by:
F = m * v^2 / r
where m is the mass of the electron and v is the speed of the orbit.
Setting these two expressions for F equal to each other, we have:
(k * |e * q|) / r^2 = m * v^2 / r
Rearranging and solving for v, we get:
v = sqrt((k * |e * q|) / (m * r))
Now, substituting the given values:
m = mass of an electron = 9.11 x 10^-31 kg
r = 0.244 nm = 0.244 x 10^-9 m
Plugging these values into the formula and evaluating, the speed of the electron in m/s is the result.
2) To calculate the magnitude of the electric field at the surface of the nucleus, we can use Gauss's law. In this case, since the charge in the nucleus is assumed to be evenly distributed, we can consider the nucleus as a uniformly charged sphere.
Gauss's law states that the electric flux through a closed surface is equal to the total charge enclosed divided by the permittivity of free space.
The electric field (E) at a point on the surface of the nucleus is given by:
E = k * (Q / r^2)
where k is the electrostatic constant, Q is the charge of the nucleus, and r is the radius of the nucleus.
Substituting the given values:
Q = charge of an electron * number of electrons in the neutral atom
r = 5.93 pm = 5.93 x 10^-12 m
Plugging these values into the formula and evaluating, the magnitude of the electric field at the surface of the nucleus in N/C is the result.
3) To find the position (x) of the third charge on the x-axis, we can use the principle of electrostatic equilibrium. In equilibrium, the net force on the third charge is zero.
The force (F1) between the first charge (q1) and the third charge (Q) is given by Coulomb's law:
F1 = (k * |q1 * Q|) / (x^2)
where k is the electrostatic constant, q1 is the charge of the first point charge, and x is the distance between the first and third charges.
The force (F2) between the second charge (q2) and the third charge (Q) is also given by Coulomb's law:
F2 = (k * |q2 * Q|) / ((14.6 cm - x)^2)
where q2 is the charge of the second point charge.
Since the net force on the third charge is zero (in equilibrium), F1 and F2 must be equal in magnitude and opposite in direction. This gives us the equation:
(k * |q1 * Q|) / (x^2) = (k * |q2 * Q|) / ((14.6 cm - x)^2)
Simplifying and solving for x, we can find the position of the third charge on the x-axis.