Question

The ionization energy of the electron in the hydrogen atom in its ground state is $13.6\, eV.$ The atoms are excited to higher energy levels to emit radiations of $6$ wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between

• A. n = 3 to n = 1 states
• B. n = 2 to n = 1 states
• C. n = 4 to n = 3 states
• D. n = 3 to n = 2 states

Answer 1

Answer: (C)

Number of spectral lines due to transition of $\vec{e}$ from nth orbit to lower orbit is $N =\frac{ n ( n -1)}{2}$ and maximum wavelength, the difference between the orbit of series should be maximum.
Number of spectral lines $N =\frac{ n ( n -1)}{2}$
$=\frac{ n ( n -1)}{2}=6 $
$\therefore n ^{2}-12=0$
$\therefore( n -4)( n +3)=0$
$n =4$
Here, $n \equiv-3$
Now as first line of series has maximum wavelength. Therefore e jump from fourth orbit to third orbit