Question

The wavelength of radiation emitted is '\(\lambda\)' when an electron jumps from second to first orbit of hydrogen atom. For the electron to jump from third to first orbit of hydrogen atom, the wavelength of radiation emitted will be

• A. \(\frac{16}{25}\lambda\)
• B. \(\frac{9}{16}\lambda\)
• C. \(\frac{27}{32}\lambda\)
• D. \(\frac{20}{27}\lambda\)

Hint:

Use the Rydberg formula for both transitions and take the ratio of the wavelengths.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The wavelength of radiation emitted in a hydrogen transition follows \(\dfrac{1}{\lambda}=R\left(\dfrac{1}{n_1^{2}}-\dfrac{1}{n_2^{2}}\right)\).

Step 2: Transition 2 to 1
\(\dfrac{1}{\lambda}=R\left(1-\dfrac{1}{4}\right)=\dfrac{3R}{4}\).

Step 3: Transition 3 to 1
\(\dfrac{1}{\lambda'}=R\left(1-\dfrac{1}{9}\right)=\dfrac{8R}{9}\).

Step 4: Calculation
\(\dfrac{\lambda'}{\lambda}=\dfrac{3/4}{8/9}=\dfrac{27}{32}\). So \(\lambda'=\dfrac{27}{32}\lambda\), which is shorter because the energy gap is larger.

Final Answer:
The new wavelength is \(\dfrac{27}{32}\lambda\). \[ \boxed{\text{(C) }\dfrac{27}{32}\lambda} \]