In a hydrogen-like sample, all the atoms are in a particular excited state. By absorbing energy of ((7/9) eV ), atoms excite to a certain state and then emit six different types of photons when they make transitions to the ground state. In the emitted photons, all the photons are having energy greater than or equal to (7/9) eV. No photon has energy smaller than this value. Then find the ionisation potential (in volts) for this H-like atom.

Question

In a hydrogen-like sample, all the atoms are in a particular excited state. By absorbing energy of ((7/9) eV ), atoms excite to a certain state and then emit six different types of photons when they make transitions to the ground state. In the emitted photons, all the photons are having energy greater than or equal to (7/9) eV. No photon has energy smaller than this value. Then find the ionisation potential (in volts) for this H-like atom. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Clearly from the given information atom must be going from

3^{rd}

to

4^{th}

state.

\frac{7}{9} = Δ E = 13.6 Z^{2} (\frac{1}{3 ^{2}} - \frac{1}{4 ^{2}})

= 13.6 z^{2} \frac{7}{144}

\Rightarrow 13.6 z^{2} = 16 e V

Clearly from the given information atom must be going from 3rd to 4th state. 97​=ΔE=13.6Z2(321​−421​) =13.6z21447​ ⇒13.6z2=16eV

Clearly from the given information atom must be going from $3^{rd}$ to $4^{th}$ state.

$\frac{7}{9} = Δ E = 13.6 Z^{2} (\frac{1}{3 ^{2}} - \frac{1}{4 ^{2}})$
$= 13.6 z^{2} \frac{7}{144}$
$\Rightarrow 13.6 z^{2} = 16 e V$