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If the wavelength of the first line in Balmer series is 6300A0, the wavelength (in A0) of the second line in the same series will be

Option: 1

6800


Option: 2

7200


Option: 3

4861


Option: 4

4700


Answers (1)

best_answer

For the first line of Balmer series

 n1 =2 , n2 = 3

\frac{1}{\lambda _{1}} = Rz^{2}\left ( \frac{1}{4}-\frac{1}{9} \right ) \Rightarrow \lambda _{1}= \frac{36}{5Rz^{2}}

For the Second line of Balmer Series :

n1 = 2, n2 = 4

\mathrm{\frac{1}{\lambda _{2}} = Rz^{2}\left ( \frac{1}{4}-\frac{1}{16} \right ) \Rightarrow \lambda _{2}= \frac{16}{3Rz^{2}}}

\frac{\lambda _{1}}{\lambda _{2}} = \frac{27}{20}

\lambda _{2} = \frac{20}{27}*\lambda _{1} = 4700A^{\circ}

 

 

Posted by

Kuldeep Maurya

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