Question

The ionization energy of the electron in the hydrogen atom in its grounds 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 = 4 to n = 3 states
• B. n = 3 to n = 2 states
• C. n = 3 to n = 1 states
• D. n = 2 to n = 1 states

Answer 1

Answer: (A)

n = 4 to n = 3 states

Number of spectral lines due to transition of $\overrightarrow{e}$ 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,$n\ne -3$

Now as first line of series has maximum wavelength. Therefore $\overrightarrow{e}$ jump from fourth orbit to third orbit