0

Guest

Courses New

ratio between longest wavelengths of H atom in lyman series to the shortest wavelengths wavelength in balmer series of He+ is?

ratio between longest wavelengths of H atom in lyman series to the shortest wavelengths wavelength in balmer series of He+ is?

Grade:11

2 Answers

jitender
114 Points
6 years ago
Wave no. = R [1/n1^2 - 1/n2^2] = 1/wavelengthTo find ratio of longest wavelength of Him Lyman and shortest of He+ in balmer series just find ratio of wave no. Of He+ to the H= R [1/2^2 - 1/~]/R [1/1^2 - 1/2^2]~= infinityLongest wave length when n2=n1+1And shortest when n2=~So ratio comes out to be 1:3
Mayur Jatinbhai Brahmbhatt
13 Points
5 years ago
1/LAMDA = RZ^2(1/22 - 1/32)
                  = R×4×(5/36)
                  = 5/9R
 
So the ratio is 5/9
This is the Right Answer
Thank You
 
Hope It Helps

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