Hydrogen atom is its ground state is excited by means of monochromatic radiation of wavelength `1023 Å`. How many different lines are possible in the resulting spectrum? Calculate the longes wavelength among them. You may assume the ionization energy of hydrogen atom as 13.6 eV.

Answer

A, C

Explanation

`E_n= E_1 = (12375)/(1023)`
or `(-13.6)/((n))^2 + 13.6 = 12.1`
Solving this equation we get n =3
`:. Total possibal emission line = (n(n-1))/(2) =3`
Longest wavelength means, minimum energy,
which is corresponding to n = 3 to n=Z.
`:. lambda_(max) = (12375)/(E_3 - E_2)`
Here, E_3 - E_2 = -(13.6)/((3))^2 + (13.6)/((4))^2 =1.9 eV`
`:. lambda_(max) = (12375)/(1.9) = 6513 Å.`