Atoms And Nuclei Question 315
Question: The ionization potential of H-atom is 13.6 V. When it is excited from ground state by monochromatic radiations of 970.6 A, the number of emission lines will be (according to Bohr’s theory)
Options:
A) 10
B) 8
C) 6
D) 4
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{1}{\lambda }=R[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} ] $
$ \Rightarrow \frac{1}{970.6\times {{10}^{-10}}}=1.097\times 10^{7}[ \frac{1}{1^{2}}-\frac{1}{n_{2}^{2}} ]\Rightarrow n_{2}=4 $
$ \therefore $ Number of emission line $ N=\frac{n( n-1 )}{2}=\frac{4\times 3}{2}=6 $
BETA
