SATHEE: Atoms And Nuclei Question 266

Atoms And Nuclei Question 266

Question: The wavelength $K_{α}$ of X-rays for two metals ‘A’ and ?B? are $\frac{4}{1875 R}$ and $\frac{1}{675 R}$ respectively, where ‘R’ is Rydberg constant. Find the number of elements lying between A and B according to their atomic numbers

Options:

A) 3

B) 1

C) 4

D) 5

Show Answer

Answer:

Correct Answer: C

Solution:

Using v $\frac{1}{λ} = R {(Z - 1)}^{2} [\frac{1}{n_{2}^{2}} - \frac{1}{n_{1}^{2}}]$ For $α$ particle, $n_{1} = 2, n_{2} = 1$

For metal A: $\frac{1875 R}{4} = R {(Z_{1} - 1)}^{2} (\frac{3}{4}) \Rightarrow Z_{1} = 26$

For metal B: $675 R = R {(Z_{2} - 1)}^{2} (\frac{3}{4}) \Rightarrow Z_{2} = 31$

Therefore, 4 elements lie between A and B.

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Atoms And Nuclei Question 266

Question: The wavelength Kα of X-rays for two metals ‘A’ and ?B? are 41875R and 1675R respectively, where ‘R’ is Rydberg constant. Find the number of elements lying between A and B according to their atomic numbers

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Question: The wavelength $K_{α}$ of X-rays for two metals ‘A’ and ?B? are $\frac{4}{1875 R}$ and $\frac{1}{675 R}$ respectively, where ‘R’ is Rydberg constant. Find the number of elements lying between A and B according to their atomic numbers