Question

In a hydrogen atom, the transition takes place from $n=3$ to $n=2$. If Rydberg's constant is $1.09\times {10}^7{m}^{-1}$, the wavelength of the line emitted is

• A. $6606\mathring{A}$
• B. $4861\mathring{A}$
• C. $4340\mathring{A}$
• D. $4101\mathring{A}$

Answer 1

The correct option is A $6606\mathring{A}$

$\frac{1}{\lambda }=(1.09\times {10}^7){\left(1\right)}^2(\frac{1}{4}-\frac{1}{9})=0.15\times {10}^7{m}^{-1}$
Therefore, wavelength emitted is $6607\mathring{A}$.