Atoms Ques 20
20. 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
[2009]
(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
Show Answer
Answer:
Correct Answer: 20.(c)
Solution:
(c) $\frac{n(n-1)}{2}=6$
$n^{2}-n-12=0$
$(n-4)(n+3)=0$ or $n=4$
If electron falls from orbit $n_2$ to $n_1$ then the number of spectral lines emitted is given by $N_E=\frac{(n_2-n_1+1)(n_2-n_1)}{2}$
If an electron falls from $n$th orbit to ground state $(n=1)$ then number of spectral lines emitted, $N_E=\frac{n(n-1)}{2}$
BETA
