×

Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

which one of the following transitions in a hydrogen atom will emit the photon of shortest wavelength n = 2 to n = 1 n = 6 to n = 5 n = 6 to n = 1 n = 5 to n = 4


2 years ago

Avni Chauhan
247 Points
							shortest wavelength implies maximum energy difference. Maximum energy difference comes when electrons are transferred from n = 2 to n = 1energy in any orbital = -13.6/ n2

2 years ago
Susmita
425 Points
							wave number$\nu =\frac{1}{\lambda} = RZ^2 (\frac{1}{n'^2}-\frac{1}{n^2})$ Shortest wavelength will correspond to largest $\nu$.For n=2 to n=1,$\nu = RZ^2 (\frac{1}{1^2}-\frac{1}{2^2})=0.75RZ^2$Similarly for n=6 to n=5, $\nu = RZ^2 (\frac{1}{5^2}-\frac{1}{6^2})=0.012RZ^2$For n=6 to n=1$\nu = RZ^2 (\frac{1}{1^2}-\frac{1}{6^2})=0.972RZ^2$For n=5 to 4$\nu = RZ^2 (\frac{1}{4^2}-\frac{1}{5^2})=0.225RZ^2$So n=6 to 1 is the answer.Please approve the answer if you are helped.

2 years ago
Think You Can Provide A Better Answer ?

Other Related Questions on General Physics

View all Questions »

Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

Course Features

• 18 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions