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 $


जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक