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)

n = 4 to n = 3 states

Answer 2

$\frac{n(n-1)}{2}=6$
$\Rightarrow n=4$
For maximum wavelength energy difference between states should be minimum because
$\lambda=\frac{ hc }{\Delta E }$
So, transition state in $n =4$ to $n =3$