# Computing energy in the electron of Li 2+?

#### robertjordan

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

2. 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...

3. 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...

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#### ehild

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

2. 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...

3. 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

"Computing energy in the electron of Li 2+?"

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