How much energy does the electron have initially in the n=4 excited state?
Enter your answer numerically in joules.
E = 2.18*10^-18/N^2
To calculate the energy of an electron in the n=4 excited state, we can use the formula for the energy levels of an electron in the hydrogen atom:
E = (-13.6 eV) / n^2
where E is the energy in electron volts (eV) and n is the principal quantum number.
First, convert the energy from electron volts to joules:
1 eV = 1.602 × 10^-19 J
Now, substitute n=4 into the equation:
E = (-13.6 eV) / (4^2) = (-13.6 eV) / 16
Convert the energy to joules:
E = (-13.6 eV / 16) * (1.602 × 10^-19 J / 1 eV)
Calculating this yields:
E = -1.09 × 10^-18 J
Therefore, the energy of the electron in the n=4 excited state is approximately -1.09 × 10^-18 J.
To calculate the energy of an electron in a specific excited state, we can use the formula:
E = -13.6 eV / n^2
where E is the energy in electron volts (eV) and n is the principal quantum number.
In this case, the electron is in the n=4 excited state.
First, we need to convert the energy from electron volts to joules. One electron volt (eV) is equal to 1.6 x 10^-19 joules.
So, the energy in joules can be calculated as:
E = -13.6 eV / (4^2) = -13.6 eV / 16
Converting to joules:
E = (-13.6 eV / 16) * (1.6 x 10^-19 J / 1 eV) = - 1.085 x 10^-19 J
Therefore, the initial energy of the electron in the n=4 excited state is approximately -1.085 x 10^-19 joules.