Answer:
The energy in its ground state is 10 meV.
Explanation:
It is given that,
The energy of the electron in its first excited state is 40 meV.
Energy of the electron in any state is given by :
For ground state, n = 1
.............(1)
For first excited state, n = 2
.............(2)
Dividing equation (1) and (2), we get :
So, the energy in its ground state is 10 meV. Hence, this is the required solution.
Answer:
3.28 cm
Explanation:
To solve this problem, you need to know that a magnetic field B perpendicular to the movement of a proton that moves at a velocity v will cause a Force F experimented by the particle that is orthogonal to both the velocity and the magnetic Field. When a particle experiments a Force orthogonal to its velocity, the path it will follow will be circular. The radius of said circle can be calculated using the expression:
r =
Where m is the mass of the particle, v is its velocity, q is its charge and B is the magnitude of the magnetic field.
The mass and charge of a proton are:
m = 1.67 * 10^-27 kg
q = 1.6 * 10^-19 C
So, we get that the radius r will be:
r = = 0.0328 m, or 3.28 cm.
