An electron is in a one-dimensional potential well of width 0.79 nm. The energy (to two decimal places and in units of eV) of this electron in its first excited state is
Question
An electron is in a one-dimensional potential well of width 0.79 nm. The energy (to two decimal places and in units of eV) of this electron in its first excited state is
Solution
The energy of an electron in a one-dimensional potential well is given by the formula:
E = n²h² / (8mL²)
where:
- E is the energy of the electron,
- n is the quantum number (n=1 for the ground state, n=2 for the first excited state, etc.),
- h is the reduced Planck's constant (h = 6.582119569 x 10^-16 eV*s),
- m is the mass of the electron (m = 9.10938356 x 10^-31 kg), and
- L is the width of the well.
For the first excited state, n=2. The width of the well, L, is given as 0.79 nm, which is 0.79 x 10^-9 m.
Substituting these values into the formula gives:
E = 2² * (6.582119569 x 10^-16 eV*s)² / (8 * 9.10938356 x 10^-31 kg * (0.79 x 10^-9 m)²)
Solving this equation gives:
E = 0.377 eV
So, the energy of the electron in its first excited state is approximately 0.38 eV.
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