Calculating Kinetic & Potential Energy of a Bohr Orbit

[DBLE]

Homework Statement

A particular Bohr orbit in a hydrogen atom has a total energy of -0.85eV. What are (i) the kinetic energy of the electron in this orbit, and (ii) the electric potential energy of teh system?

Homework Equations

E=mc^2

KE= 1/2mv^2

E = hc/lambda

The Attempt at a Solution

I'm a little confused in with this question. Wouldn't the KE be 0.85eV? and is the Potential energy given with mc^2?

really appreciate any help given =)

 

[mark726]

Hello! I can help clarify this for you. In this scenario, the total energy of the electron in the Bohr orbit is given as -0.85eV. This means that the sum of the kinetic energy and potential energy of the system is -0.85eV. So, to find the kinetic energy, we can rearrange the equation to solve for KE:

KE = E - PE

KE = -0.85eV - PE

We do not have enough information to directly solve for the potential energy, but we can use the equations you listed to find it indirectly. The first equation, E=mc^2, is the famous mass-energy equivalence equation. This equation tells us that the energy of a particle is equal to its mass multiplied by the speed of light squared. However, this equation is typically used for massive particles, so we can use a modified version for an electron:

E = (m_e)(c^2)

Where m_e is the mass of the electron. The second equation, KE = 1/2mv^2, is the classical equation for kinetic energy. This can be used to find the velocity of the electron in the Bohr orbit, since we know its mass (m_e) and we know its total energy (-0.85eV). So, we can set up an equation to solve for v:

KE = (1/2)(m_e)(v^2)

-0.85eV = (1/2)(m_e)(v^2)

v = √(2(-0.85eV)/(m_e))

Now that we know the velocity of the electron in the Bohr orbit, we can use the third equation, E=hc/λ, to find the potential energy. This equation relates the energy of a particle to its wavelength (λ). Since we know the energy (-0.85eV) and we can find the wavelength using the velocity we just calculated, we can solve for the potential energy:

-0.85eV = (hc)/λ

λ = (hc)/(-0.85eV)

Now that we know λ, we can plug this into the equation for potential energy:

PE = (hc)/λ

PE = (hc)/[(hc)/(-0.85eV)]

PE = -0.85eV

So, the potential energy of the system is also -0.85eV. This makes