0

Guest

Courses New

H atom is exposed to electromagnetic radiation of 1028 angstrom and gives out induced radiations .calculate lamda of induced radiations.

H atom is exposed to electromagnetic radiation of 1028 angstrom and gives out induced radiations .calculate lamda of induced radiations.

Grade:11

1 Answers

Vikas TU
14149 Points
6 years ago
Dear Student,
 As we realize that, 1 Joule = 6.24 x 1018 eV, at that point 1.93 x 10-18 Joule = (1.93 x 10-18 ) x (6.24 x 1018 ) = 12.06 eV 
Since, H-molecule in its ground state has the vitality of 13.6 eV, at that point: 
The vitality required to take the electron in the ground state to first energized state is:
E1 - E2= 13.6 - (13.6/n2) = 13.6 - (13.6/22) = 13.6 - 3.4 = 10.2 eV 
So also, the vitality required to take the electron in the ground state to second energized state is: E1 - E3 = 13.6 - (13.6/32) = 13.6 - 1.51 = 12.08 eV 
Accordingly, the incited radiation from this state will be: 
λ1 = n = 3 - > n = 1 ; λ2 = n = 3 - > n = 2 ; λ3 = n = 2 - > n = 1 
λ1=hc∆E=hcE1−E36.626×10−34 J s×3×108 ms−1(13.6 − 1.51)×1.6×10−19 J=1.027×10−7 m = 1027  Aoλ3=hc∆E
=hcE1−E26.626×10−34 J s×3×108 ms−1(13.6 − 3.4)×1.6×10−19 J
=1218 A.
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

 

Think You Can Provide A Better Answer ?

Other Related Questions on Physical Chemistry

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More