The binding energy of a deuteron is `2times10^(6)eV` .Find the minimum kinetic energy a proton should possess to split the deuteron (in `10^(6)eV`).

If ionising energy of H atom is 13.6 eV, then the second ionising energy of He should be

The kinetic energy of an electron in hydrogen atom is 3.40 eV .The minimum energy required to ionise the hydrogen atom is

If the energy of a photon is 10 eV , then its momentum is

The binding energy of an electron in the ground state of He is equal to 24.6 eV: Find the energy required to remove both electrons.

The energy of the electron in the ground state of hydrogen atom is -13.6 eV . Find the kinetic energy and potential energy of electron in this state.

if the binding energy per nucleon of deuteron is 1.115 Me V , find its masss defect in atomic mass unit .

Photons of energy 6 eV are incident on a metal surface whose work function is 4 eV . The minimum kinetic energy of the emitted photo - electrons will be

" The binding energy for the third electron in the ground state of "Li" -atom (in eV) should be "

Calculate the binding energy of the deuteron, which consistss of a proton and a neutron, given that the atomic mass of the deuteron is 2.014102 amu,. Take mass of proton (m_(p))=1.007825 amu, mass of a neutron (m_(n))=1.008665 amu and 1amu=931.5 MeV