Atoms And Nuclei Question 319

Question: Taking Rydberg’s constant $ R=1.097\times 10^{7}m $ , first and second wavelength of Balmer series in hydrogen spectrum is

Options:

A) $ 2000\overset{o}{\mathop{\text{A }~}},,3000\overset{o}{\mathop{\text{A }~}},~ $

B) $ \text{1575 }\overset{o}{\mathop{\text{A }~}},\text{, }~\text{ 2960 }\overset{o}{\mathop{\text{A }~}}, $

C) $ \text{6529 }\overset{o}{\mathop{A}},\text{, },\text{4280 }\overset{o}{\mathop{\text{A }~}}, $

D) $ 6552\overset{o}{\mathop{\text{A }~}},,4863\overset{o}{\mathop{\text{A }~}},~~ $

Show Answer

Answer:

Correct Answer: D

Solution:

  • $ \frac{1}{\lambda }=R[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} ]. $ For first wavelength, $ n_{1} $ $ =2,n_{2}=3 $
    $ \Rightarrow {\lambda_{1}}=6563\overset{o}{\mathop{A}},. $ For second wavelength, n, $ =2,n_{2}=4 $
    $ \Rightarrow {\lambda_{2}}=4861\overset{o}{\mathop{A}}, $


जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक