An electron is in excited state in a hydrogen-like atom. It has a total energy of `-3.4eV` .The kinetic energy of the electron is E and its de Broglie wavelength is `lambda`

In an excited state of hydrogen like atom an electron has total energy of -3.4 eV . If the kinetic energy of the electron is E and its de-Broglie wavelength is lambda , then

The kinetic energy of an electron with de - Broglie wavelength of 0.3 nanometre is

An electron in a hydrogen like atom is in an excited state3 . It has a total energy of -3 4 eV. Calculate : (a) The kinetic energy of electron . (b) The de-Broglie wavelength of electron . ( h = 6.6 xx 10^(023) , m_e = 9 .108 xx 10^(-31) kg)

An electron , in a hydrogen-like atom, is in excited state. It has a total energy of -3.4 eV. Calculate (a) the kinetic energy and (b) the de Broglie wavelength of the electron.

An electron, in a hydrogen like atom , is in excited state. It has a total energy of -3.4 eV, find the de-Broglie wavelength of the electron.

Total energy of electron in an excited state of hydrogen atom is -3.4 eV . The kinetic and potential energy of electron in this state.