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The wavelength K_{\alpha } 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

Option: 1

3


Option: 2

1


Option: 3

4


Option: 4

5


Answers (1)

best_answer

\text { Using } \frac{1}{\lambda}=R(z-1)^2\left[\frac{1}{\mathrm{n}_2^2}-\frac{1}{\mathrm{n}_1^2}\right]

For \alpha -particle, n_{1}=2, n_{2}=1

For\; metal \; A: \frac{1875 \mathrm{R}}{4}=\left(Z_1-1\right)^2\left(\frac{3}{4}\right) \Rightarrow Z_1=31

For \; metal\; \mathrm{B}: 675 \mathrm{R}=\mathrm{R}\left(\mathrm{Z}_2-1\right)^2\left(\frac{3}{4}\right) \Rightarrow \mathrm{Z}_2=31

Therefore, 4 elements lie between A and B. = 220MeV.

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Pankaj

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