A particle in the first excited state of a one-dimensional infinite potential energy well (with U = 0 inside the well) has an energy of 6.0 eV. What is the energy of this particle in the ground state?

The energy of this particle in the ground state is E₁=1.5 eV.

Explanation:

The energy of a particle of mass m in the nth energy state of an infinite square well potential with width L is:

In the ground state (n=1). In the first excited state (n=2) we are told the energy is E₂= 6.0 eV. If we replace in the above equation we get that:

So we can rewrite the energy in the ground state as:

Finally

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft

the components of a vector are the parts of the vectors that are to each other.

b perpendicular