Question

A hydrogen like atom, of atomic number z is in its (2n)^th state. From this state, the atom can emit a photon of maximum energy 204 eV. It makes a transction, from this excited state, to a state of quantum number n, emitting a photon of energy 40.8 eV. For the atom

(1) z = 4; n = 2

(2) z = 3; n = 3

(3) z = 4; n = 4

(4) z = 3; n = 4

Answer 1

(1) z = 4; n = 2

Let E₁ be the energy of the atom in its ground state. For atom in (2n)^th state E_2n = [E₁/(2n)²].

The maximum energy photon is emitted when electron jumps from this state to the ground state, i.e.,

When electron jumps, from n₂ (= 2n) to n₁ (= n) state, the energy of emitted photon is 40.8 eV. Therefore

From Eqns. (1) and (2), we have

Substituting this value of n, in Eqn. (2), we get