Question

The ionization potential of hydrogen is 13.6 V. Calculate the energy of the
radiations emitted corresponding to a transition from n =3 to n = 2 level.​

Answer 1

Answer:n=3

Explanation:

Ionization energy corresponding to ionization potential =−13.6eV

Photon energy incident =12.1eV

So, the energy of electron in excited state

=−13.6+12.1=−1.5eV

ie, E

n

​

=−

n

2

13.6

​

eV

−1.5=

n

2

−13.6

​

⇒ n

2

=

−1.5

−13.6

​

≈9

∴ n=3

ie, energy of electron in excited state corresponds to third orbit.

The possible spectral line are when electron jumps from orbit 3rd to 2nd; 3rd to 1st and 2nd to 1st. Thus, 3 spectral lines are emitted.