Question

The wavelengths of K_a X-rays of the metals 'A' and 'B' are $\frac{4}{1875R}$ and $\frac{1}{675R}$respectively. Where 'R' is rydber constant. The number of elements lying between 'A' and 'B' according to their atomic numbers, is-

• A. 3
• B. 6
• C. 5
• D. 4

Answer 1

Answer: (D)

4

Answer 2

Using $\frac { 1 } { \lambda }$ = R (z – 1)²

for a particle ; n₁ = 2, n₂ = 1

for metal A ; $\frac { 1875 R } { 4 }$ = R (Z₁ – 1)² $\left[ \frac { 3 } { 4 } \right]$Ž Z₁ = 26

for metal B ; 675 R = R (Z₂ – 1)² $\left[ \frac { 3 } { 4 } \right]$Ž Z₂ = 39

Therefore, 4 elements lie between A and B.