In Hydrogen atom, energy of first excited state is -3.4 eV. Then find out KE of same orbit of Hydrogen atom :

NEET 2002 Physics question

In Hydrogen atom, energy of first excited state is eV. Then find out KE of same orbit of Hydrogen atom :

Answer: (1)

In a Bohr orbit the electron's kinetic energy is , potential energy and total energy . Hence . For the first excited state () $E = -3.4\

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Question ID VK-088986