428 Views

What is the frequency and wavelength photon emmited during a transition from n=5 n=2 from an hydrogen atom


AKASH A 9th Aug, 2020
Answer (1)
Ankita Grover 9th Aug, 2020

Using Rydberg’s formula, we have: v = R*[1/(n1^2) – 1/(n2^2)]

where v = frequency in cm^-1

R = Rydberg’s Constant = 109677.6 in cm^-1

n1 = 2 (Given)

n2 = 5 (Given)

Substituting the values, we get:

v = 109677.6 * [1/(2^2) – 1/(5^2)]

V = 109677.6 * [1/4-1/25]

v = 23,032.296 cm^-1 (Answer)

We know, wavelength = 1/frequency = 1/23032.296

Wavelength = 4.34 * 10^-5 cm (Answer)

Related Questions

MAHE Manipal M.Tech 2025
Apply
NAAC A++ Accredited | Accorded institution of Eminence by Govt. of India | NIRF Rank #4
Great Lakes PGPM & PGDM 2025
Apply
Admissions Open | Globally Recognized by AACSB (US) & AMBA (UK) | 17.3 LPA Avg. CTC for PGPM 2024 | Application Deadline: 10th March 2025
Amity University, Noida Law A...
Apply
700+ Campus placements at top national and global law firms, corporates, and judiciaries
Chandigarh University Admissi...
Apply
Ranked #1 Among all Private Indian Universities in QS Asia Rankings 2025 | Scholarships worth 210 CR
Amrita Vishwa Vidyapeetham | ...
Apply
Recognized as Institute of Eminence by Govt. of India | NAAC ‘A++’ Grade | Upto 75% Scholarships
UPES B.Tech Admissions 2025
Apply
Ranked #42 among Engineering colleges in India by NIRF | Highest CTC 50 LPA , 100% Placements | Last Date to Apply: 28th March
View All Application Forms

Download the Careers360 App on your Android phone

Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile

150M+ Students
30,000+ Colleges
500+ Exams
1500+ E-books