An electron makes transition from n=3 to the orbit n=2 of a hydrogen atom (ionization potential 13.6 eV ). The energy of the photon emitted in the process is

Question

An electron makes transition from n=3 to the orbit n=2 of a hydrogen atom (ionization potential 13.6 eV ). The energy of the photon emitted in the process is (A) 1.89 eV (B) 2.55 eV (C) 12.09 eV (D) 12.75 eV

Tardigrade Admin · Accepted Answer

Energy released =13.6[ (1/(1)2)-(1/(3)2) ] =12.09 text eV

Q. An electron makes transition from $n = 3$ to the orbit $n = 2$ of a hydrogen atom (ionization potential $13.6 e V$ ). The energy of the photon emitted in the process is

1.89 eV

2.55 eV

12.09 eV

12.75 eV

Energy released
$= 13.6 [\frac{1}{( 1 ) ^{2}} - \frac{1}{( 3 ) ^{2}}]$
$= 12.09 e V$

Q. An electron makes transition from n=3 to the orbit n=2 of a hydrogen atom (ionization potential 13.6eV ). The energy of the photon emitted in the process is