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

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 a total energy of -3.4 eV. Then its de Broglie wavelength is nearly

An electron in a hydrogen -like atom is in an excited state. It has a total energy of -3.4 eV. Calculate the kinetic energy of the electron.

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 is having a kinetic energy of 50 eV. Its de-Broglie wavelength is

The kinetic energy of an electron is 500 eV. Calculate the de-Broglie wavelength associated with it.