The de-Broglie's wavelength of an electron in the 4th orbit is _________ .

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
The de-Broglie's wavelength of an electron in the 4th orbit is _________ $π a_{0}$ . ( $a_{0}$ = Bohr's radius) [2024]

Ans.
(8)   De-Broglie's wavelength ( $λ$ ) is related to mass (m) and velocity v of electron by:

   $λ = \frac{h}{m v} - I,$ where $h$ is Planck's constant.

By quantization of angular momentum:

$m v r = \frac{n h}{2 π} - II$

Where $n$ is orbit number and $r$ is radius of $n^{t h}$ orbit.

Rearranging equation II

$2 π r = n \frac{h}{m v} - III$

From I and III

   $2 π r = n λ - IV$

Radius ( $r$ ) of $n^{t h}$ orbit is related to radius of first orbit ( $a_{0}$ ) as:

$r = a_{0} \frac{n^{2}}{Z} - V$

From IV and V:

$2 π a_{0} \frac{n^{2}}{Z} = n λ$

$λ = 2 π a_{0} \frac{n}{Z}$

Taking atom to be H, Z = 1 and for fourth orbit, n = 4

   $λ = 2 π a_{0} \frac{4}{1} = 8 π a_{0}$

Want Us to Email you About Special Offers & Updates?