Determine the value of the de Broglie wavelength associated with the electron orbiting in the ground state of hydrogen atom. How will the de Broglie wavelength change when it is in the first excited states.

Open in App

Solution

Dear Student

$T h e e n r g y o f n t h s t a t e o f h y d r o g e n a t o m i s g i v e n b y E_{n} = - \frac{13.6}{n^{2}} S o i n g r o u n d s t a t e n = 1 E_{1} = - 13.6 e V T h e r e f o r e λ = \frac{h c}{E} = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{13.6 \times 1.6 \times 10^{- 19} J} = 914.0 A W h e n a t o m i s i n f i r s t e x c i t e d s t a t e n = 2 S o e n e r g y E_{2} = - \frac{13.6}{4} e V = - 3.4 e V C h a n g e i n e n e r g y = E_{2} - E_{1} = - 3.4 e V + 13.6 e V = 10.2 e V T h e r e f o r e c h a n g e i n w a v e l e n g t h ∆ λ = \frac{h c}{∆ E} = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{10.2 \times 1.6 \times 10^{- 19} J} = 1218.7 A$

Suggest Corrections

Similar questions

Q. The de-Broglie wavelength

(λ_{B})

associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state

(λ_{G})

by :

Q. When an electron in hydrogen atom jumps from the third excited state to the ground state. how would the de broglie wavelength have associated with the electron change?

Given the ground state energy $E_{0} = - 13.6 e V$ and Bohr radius $a_{0}$ = 0.53 A . Find out how the de Broglie wavelength associated with the electron orbiting in the ground state would change when it jumps into the first excited state.

Q. When an electron in hydrogen atom jumps from the third excited state to the ground state, how would the de-Broglie wave length associated with the electron change? Justify your answer.

Q. The De-Broglie wavelength associated with electrons revolving round the nucleus in a hydrogen atom in ground state, will be-

Join BYJU'S Learning Program