Atoms And Nuclei Question 326
Question: The third line of the Balmer series spectrum of a hydrogen like ion of atomic number Z equals to 108.5 nm. Then Z is
Options:
A) 2
B) 5
C) 3
D) 6
Show Answer
Answer:
Correct Answer: A
Solution:
- For the third line of Balmer series,
$ n_{1}=2,n_{2}=5 $
$ \therefore \frac{1}{\lambda }=RZ^{2}( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} )=RZ^{2}( \frac{1}{2^{2}}-\frac{1}{5^{2}} )=\frac{21RZ^{2}}{100} $
$ E=-13.6eV $ $ Z^{2}\times \frac{21}{100}=\frac{hc}{\lambda }=\frac{1242eVnm}{108.5nm} $
$ Z^{2}=\frac{1242\times 100}{108.5\times 21\times 13.6}=4\Rightarrow Z=2 $
BETA
