Consider an electron (m_e = 9.1 × 10^-31 kg) having energy 13.6 eV, confined in an infinite potential well. If the potential energy inside the well is zero, the expectation value for the square of the electron speed, (v²), is

Concept:

• Average Energy/ Expectation value of energy of an electron is the sum of expectation value of kinetics and potential energy and it is written as

\(<H>=<T>+<V> \)

\(where,\\<H>= Total\;energy\)

\(<T>= Avg.\;kinetic \;energy\)

\(<V>= Avg.\;potential\;energy\)

• Also, expectation value of kinetic energy of electron is related to mass of electron (m_e) and the velocity (v) as:

\(<T>=\frac{1}{2}m_e<v^2>\)

Explanation:

According to the question, the potential energy of the particle/electron inside the well is 0. So, it can be concluded the given energy value is the kinetic energy of the electron

\(<H>=\frac{1}{2}m_e<v^2>+0\)

\(<H>=\frac{1}{2}m_e<v^2>\)

The above relation can be rearranged to get the expectation value of square of velocity as follows:

\(<v^2>=\frac{2<H>}{m_e}\;\)       ......relation(1)

given,

\(<T>=13.6eV=13.6\times1.6\times10^{-19}J \)

\(m_e=9.1\times10^{-31}Kg \)

Equating the values in relation(1), will give:

\(<v^2>= 2\times\frac{13.6\times1.6\times10^{-19}J}{9.1\times10^{-31}Kg}\;\)

\(<v^2>=4.7\times10^{12}m^2s^{-2} \)

Conclusion:

Hence, the expectation value of square of velocity of electron confined in the infinite potential well is \(4.7\times10^{12}m^2s^{-2} \)