A) 3
B) 1
C) 4
D) 5
Correct Answer: C
Solution :
[c] Using v\[\frac{1}{\lambda }=R{{\left( Z-1 \right)}^{2}}\left[ \frac{1}{n_{2}^{2}}-\frac{1}{n_{1}^{2}} \right]\] For \[\alpha \] particle, \[{{n}_{1}}=2,{{n}_{2}}=1\] 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: \[675R=R{{\left( {{Z}_{2}}-1 \right)}^{2}}\left( \frac{3}{4} \right)\Rightarrow {{Z}_{2}}=31\] Therefore, 4 elements lie between A and B.
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