Computing energy in the electron of Li 2+?

[robertjordan]

Homework Statement

Using the Bohr model of the atom, compute the energy in eV of the one electron in Li²⁺.

Homework Equations

E_n=\frac{m\cdot e^4 \cdot z^2}{2n^2 \cdot \hbar^2}
Where m= mass of electron, z= atomic number, e= charge of an electron, n is the energy level.

^ I think those are the right meanings of the variables...

The Attempt at a Solution

plugging in 1 for n, 3 for z, 9.11*10^-31 kg for m, -1.602*10^-19 Coulombs for e, and 1.054572×10^-34 J*s for h, we get
E_1= ((9.11*10^-31 kg))*((-1.602*10^-19 C)^4)*(3^2)/(2(1.054572×10^-34 J s)^2) = 2.428×10^-37 s^6A^4/(kg m^4) (second to the 6 amperes to the fourth per kilogram meter to the fourth).

This clearly doesn't seem right. What am I doing wrong? I need help...

 

[ehild]

Homework Statement

Using the Bohr model of the atom, compute the energy in eV of the one electron in Li²⁺.

Homework Equations

E_n=\frac{m\cdot e^4 \cdot z^2}{2n^2 \cdot \hbar^2}
Where m= mass of electron, z= atomic number, e= charge of an electron, n is the energy level.

^ I think those are the right meanings of the variables...

The Attempt at a Solution

plugging in 1 for n, 3 for z, 9.11*10^-31 kg for m, -1.602*10^-19 Coulombs for e, and 1.054572×10^-34 J*s for h, we get
E_1= ((9.11*10^-31 kg))*((-1.602*10^-19 C)^4)*(3^2)/(2(1.054572×10^-34 J s)^2) = 2.428×10^-37 s^6A^4/(kg m^4) (second to the 6 amperes to the fourth per kilogram meter to the fourth).

This clearly doesn't seem right. What am I doing wrong? I need help...

Check the equation for the energy. ε₀² is missing from the denominator.

E_n=\frac{m\cdot e^4 \cdot z^2}{2n^2 \cdot \hbar^2\cdot ε_0^2}

ehild