Question

In a sample of h atoms electrons de excite from n to 1. total number of lines belonging to the balmer series are two. if electrons are ionised from n by photons of energy 13 EV then the kinetic energy of the ejected photoelectrons will be

12.15ev
11.49ev
12.46ev
12.63ev​

Answer 1

in Balmer series , electron transitions upto second energy level. I mean, $n_1=2$ and $n_2=n$

a/c to question, total number of lines belonging to the balmer series are two.

mean, This corresponds to transitions 4→2 , 3→2

so, n = 4

now, energy of 4th level for H-atom is, E = 13.6/(4)² = 0.85 eV

given, electrons are ionized from n level (n = 4) by photons of energy 13 eV

so, kinetic energy of ejected photoelectrons will be = 13 eV - E

= 13 eV - 0.85 eV

= 12.15 eV

hence, option (1) is correct choice.

Answer 2

Answer:

Balmer series , electron transitions upto second energy level. I mean, n_1=2n1=2and n_2=nn2=n

a/c to question, total number of lines belonging to the balmer series are two.

mean, This corresponds to transitions 4→2 , 3→2

so, n = 4

now, energy of 4th level for H-atom is, E = 13.6/(4)² = 0.85 eV

given, electrons are ionized from n level (n = 4) by photons of energy 13 eV

so, kinetic energy of ejected photoelectrons will be = 13 eV - E

= 13 eV - 0.85 eV

= 12.15 eV