Estimate the ratio of the wavelengths associated with the electron orbiting around the nucleus in the first and second excited states of hydrogen atom.

 

 
 
 
 
 

Answers (1)
S safeer

The wavelength can find by using De-Broglie wavelength equation,

for electron, the first excited n=2.

By de-broglie's equation, n\lambda =2\pi r _{n}

where, r _{n} is the radius of second orbit,

r _{n}=n^{2}\left ( \frac{h^{2}\epsilon _{0}}{\pi m_{e}^{2}} \right )

      =n^{2}\times 0.527\; A^{\circ}

      =2.11\; A^{\circ}

Therefore, 2\times \lambda =2\times 3.14\times 2.11

                    \lambda =6.625\; A^{\circ}

For second excited state, n=3

            So, n\lambda =2\pi r_{n}

                    r_{n}=n^{2}\times 0.529\; A^{\circ}

                           =9\times 0.529\; A^{\circ}=4.761\; A^{\circ}

Therefore, 3\times \lambda =2\times 3.14\times 4.761

                        \lambda =\frac{29.899}{3}=9.966\; A^{\circ}

    Ratio,               =\frac{6.625}{9.966}

                            =0.66

ratio of the wave lengths is 0.66

Preparation Products

Knockout NEET July 2020

An exhaustive E-learning program for the complete preparation of NEET..

₹ 15999/- ₹ 6999/-
Buy Now
Rank Booster NEET 2020

This course will help student to be better prepared and study in the right direction for NEET..

₹ 9999/- ₹ 4999/-
Buy Now
Knockout JEE Main July 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
Buy Now
Test Series NEET July 2020

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 11999/-
Buy Now
Exams
Articles
Questions