The radius of hydrogen atom in its ground state is `5.3 xx 10^(-11)m`. After collision with an electron it is found to have a radius of `21.2 xx 10^(-11)m`. What is the principle quantum number of `n` of the final state of the atom ?
The radius of hydrogen atom in its ground state is 5.3 xx 10^(-11)m . After collision with an electron, it is found to have a radius of 21.2 xx 10^(-11)m . What is the principal quantum number of the final state of the atom?
The radius of hydrogen atom in its ground state is 5.3 xx 10^-11 m . After collision with an electron it is found to have a radius of 21.2 xx 10^-11 m . The principal quantum number of the final state of the atom is.
The radius of hydrogen atom, in its ground state, is of the order of
The radius of H atom in its ground state is 0.53 Å After collision with an electron, its radius is found to be 21.2 xx 10^(-11)m . In this state, the principle quantum number 'n' of the H-atom is
The radius of hydrogen atom in the ground state is 5.3 xx 10^(-11) m. When struck by an electron, its radius is found to be 21.2 xx 10^(-11) m . The principal quantum number of the final state will be
The radius of the hydrogen atom in its ground state is 5.3xx10^(-11) m. The principal quantum number of the final state of the atom is
In the ground state of hydrogen atom, its Bohr radius is 5.3xx10^(-11)m . The atom is excited such that the radius becomes 21.2xx10^(-11)m . Find the value of principal quantum number and total energy of the atom in excited state.
