Question

Deuterium atoms in the ground state, are radiated by photons of energy $12.8eV$. What will be the energy of induced radiation of longest wavelength? (Ionization energy of deuterium is $14.4eV$)

• A. $12.8eV$
• B. $10.8eV$
• C. $1.6eV$
• D. $2.00eV$

Answer 1

The correct option is D $2.00eV$

Energy of photon $=12.8eV$
If the photon fully loses its energy, the orbit to which electron from ground state will reach is given as,
$\frac{14.4}{n^2}=(14.4\to 12.8)$
$\frac{14.4}{1.8}=n^2\Rightarrow n^2=09\Rightarrow n\approx 3$

For longest wavelength or smallest frequency, transition should be made from $n=3$ to $n=2$ and the energy released is in form of photon.

So, energy released $=14.4[\frac{1}{2^2}-\frac{1}{3^2}]=14.4\left[\frac{9-4}{36}\right]eV$$=2eV$