Question

The wavelength of $L_{\alpha }$ line from an element of atomic number 78 is ‘$\lambda$ ’ If the atomic number is $48$ the corresponding wavelength of $L_{\alpha }$ line?
($b=6.4$ for $L_{\alpha }$ lines)

• A. $1.98$$\lambda$
• B. $2.96$$\lambda$
• C. $3.42$$\lambda$
• D. $8.3$$\lambda$

Answer 1

Answer: (B)

$2.96$$\lambda$

$\frac{1}{\lambda }=R(2-6{)}^2)(\frac{1}{n_{1^2}}-\frac{1}{n_{2^2}})$
$\frac{1}{\lambda }=R(78-6.4{)}^2\left(\frac{5^1}{36}\right)$
$\lambda -\frac{1}{{\lambda }^1}=R(48-6.4{)}^2\left(\frac{5}{36}\right)$
$\frac{{\lambda }^1}{\lambda }=3$
${\lambda }^1=3\lambda$