Atoms And Nuclei Question 316

Question: The difference between the longest wavelength line of the Balmer series and shortest wavelength line of the Lyman series for a hydrogenic atom (atomic number Z) equal to $ \Delta \lambda $ . The value of the Rydberg constant for the given atom is :

Options:

A) $ \frac{5}{31}\frac{1}{\Delta \lambda .Z^{2}} $

B) $ \frac{5}{36}\frac{Z^{2}}{\Delta \lambda .} $

C) $ \frac{31}{5}\frac{1}{\Delta \lambda .Z^{2}} $

D) none of these

Show Answer

Answer:

Correct Answer: C

Solution:

  • $ \Delta \lambda ={\lambda_{1}}-{\lambda_{2}} $ $ \frac{1}{R[ \frac{1}{{{( 2 )}^{2}}}-\frac{1}{{{( 3 )}^{2}}} ]Z^{2}}=-\frac{1}{R[ 1-0 ]Z^{2}} $ . On solving, $ R=\frac{31}{5}\frac{1}{\Delta \lambda .Z^{2}} $


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