Question

Calculate the wavelength of radiation emitted when He+ makes a transition from the state n = 3 to the state n = 2
1. 180 nm
2. 164 nm
3. 464 nm
4. 506 nm

Answer 1

Correct Answer - Option 2 : 164 nm

When an electron jumps from mth orbit to the nth orbit (m>n), the energy of the atom changes from E_m to E_n. This extra energy (E_m - E_n) is emitted as a photon of electromagnetic radiation. This corresponding wavelength is given as

\(\frac{1}{\lambda }\; = \;R{Z^2}\left( {\frac{1}{{{n^2}}} - \frac{1}{{{m^2}}}} \right)\)

Where R is the Rydberg constant = 1.0973 × 107 m - 1

Putting values z = 2, for He+ ion.

\(\begin{array}{l} \frac{1}{\lambda }\; = \;4R\left( {\frac{1}{4} - \frac{1}{{9}}} \right)\; = \;4R \times \frac{5}{{36}}\; = \;\frac{5}{{9}}R\\ \lambda \; = \;\frac{{9}}{{5R}}\; = \;\frac{{9}}{{5 \times 1.0973 \times {{10}^7}}}\; = \;164\;nm \end{array}\)