Question

In a sample of $H-$atoms,elelctrons de-excite from a level $n$ to $1$.The total number of lines belonging to Balmer series are two. If the elelctrons are ionised from level n by photons of energy $13\text{eV}$. Then the kinetic energy of the ejected photoelelctrons will be:

• A. $12.15\text{eV}$
• B. $11.49\text{eV}$
• C. $12.46\text{eV}$
• D. $12.63\text{eV}$

Answer 1

The correct option is A $12.15\text{eV}$

$\text{Number of lines in Balmer series =2}$
$\therefore n=4(\text{lines will be}4\to 2,3\to 2)$.
Kinetic energy of ejected photoelelctrons $=E_{photon}-B{E}_n=13-\frac{13.6}{4^2}$
$=13-0.85=12.15\text{eV}$