Question 317: Dual Nature of Matter and Radiation

Question

When the electron orbiting in hydrogen atom in its ground state moves to the third excited state, the de-Broglie wavelength associated with it

Accepted Answer

To determine how the de Broglie wavelength of an electron changes when it transitions from the ground state to the third excited state in a hydrogen atom, we need to analyze the relationship between the de Broglie wavelength and the principal quantum number $ n $.
Principal Quantum Number and Energy Levels:
Ground State:
$ n = 1 $
First Excited State:
$ n = 2 $
Second Excited State:
$ n = 3 $
Third Excited State:
$ n = 4 $
So, the electron transitions from $ n = 1 $ to $ n = 4 $.
De Broglie Wavelength in the Bohr Model:
In the Bohr model, the de Broglie wavelength $ \lambda_n $ of an electron in the $ n^	ext{th} $ orbit is given by:
$ \lambda_n = \frac{h}{m v_n} $
Where:
$ h $ is Planck's constant.
$ m $ is the mass of the electron.
$ v_n $ is the velocity of the electron in the $ n^	ext{th} $ orbit.
Velocity of the Electron:
The velocity $ v_n $ is inversely proportional to $ n $:
$ v_n = \frac{v_1}{n} $
Substituting $ v_n $ into $ \lambda_n $:
$ \lambda_n = \frac{h}{m v_n} = \frac{h}{m \left( \frac{v_1}{n} ight)} = n \left( \frac{h}{m v_1} ight) = n \lambda_1 $
This shows that the de Broglie wavelength is directly proportional to the principal quantum number $ n $:
$ \lambda_n = n \lambda_1 $
Conclusion:
As the electron moves from the ground state ($ n = 1 $) to the third excited state ($ n = 4 $), the de Broglie wavelength increases by a factor of 4:
$ \lambda_4 = 4 \lambda_1 $
Answer: Option D Ã¢ÂÂ The de Broglie wavelength will increase.
Answer:
Will increase.

Answer

becomes zero.

Answer

remains unchanged.

Answer

will decrease.

Answer

will increase.

Dual Nature of Matter and Radiation

When the electron orbiting in hydrogen atom in its ground state moves to the third excited state, the de-Broglie wavelength associated with it

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