Question

The wavelength of light emitted by electron in $C^{5+}$ ion during transition from $n=3$ to $n=1$ is approximately:

• A. $2.85nm$
• B. $3.6nm$
• C. $8.7nm$
• D. $9.8nm$

Answer 1

Answer: (A)

$2.85nm$

We know that
$\frac{1}{\lambda }=Z^{2}R(\frac{1}{h_1^2}-\frac{1}{h_1^2})$
$=-(6{)}^2\times 1.097\times {10}^7(\frac{1}{(3{)}^2}-\frac{1}{(1{)}^2})$
$\frac{1}{\lambda }=39.492\times \frac{1-9}{9}$
$\frac{1}{\lambda }=\frac{315.93\times {10}^7}{9}=35.104{\times }^7$
$\lambda =\frac{1}{35.104\times {10}^7}=0.0285\times {10}^{-7}$
$\lambda =2.85nm$