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The wavelengths of Ka x-rays of two metals ‘A’ and ‘B’ are  \frac{4}{1875 R} and \frac{4}{675 R} respectively, where ‘R’ is Rydberg's constant. The number of elements lying between ‘A’ and ‘B’ according to their atomic numbers is

Option: 1 3

Option: 2 6

Option: 3 5

Option: 4 4

Answers (1)

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 According to Moseley's equation 

\frac{1}{\lambda }=R\left ( z-1 \right )^{2}\left [ \frac{1}{n_{2}^{2}} - \frac{1}{n_{1}^{2}} \right ]

 For k_{\alpha} particle; n1 =2,n2 =1

\frac{1}{\lambda }=R\left ( z-1 \right )^{2}\left [ \frac{1}{1^{2}} - \frac{1}{2^{2}} \right ]

 

For metal A ; \frac{1875R}{4}=R\left ( Z_{1}-1 \right )^{2}\left ( \frac{3}{4} \right )

\Rightarrow z_{1}=26

For metal B;  \frac{675R}{4}=R\left ( Z_{2}-1 \right )^{2}\left ( \frac{3}{4} \right )\Rightarrow z_{2}=31

Therefore,  4 elements lie between A and B, i.e. with Z= 27, 28 ,29 , 30

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Divya Prakash Singh

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