⇦ | ![]() | ⇨ |
An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be: (m is the mass of electron, R, Rydberg constant and h Planck’s constant)
Options
(a) 24hR / 25m
(b) 25hR / 24m
(c) 25m / 24hR
(d) 24m / 25hR
Correct Answer:
24hR / 25m
Explanation:
For emission, the wave no. of the radiation is given as
1/λ = Rz² (1/n₁² – 1/n₂²)
R = Rydberg constant, Z = atomic number
= R (1/1² – 1/5²) = R (1 – 1/25) ⇒ 1/λ = R 24/25
linear momentum
P = h/λ = h x R x 24/25 (de-Broglie hypothesis)
⇒ mv = 24hR / 25 ⇒ V = 24hR / 25m
Related Questions:
- The threshold frequency for a photosensitive metal is 3.3 x 10¹⁴ Hz.
- In a parallel plate capacitor with plate area A and charge Q, the force on one plate
- An aeroplane flies 400m due to North and then 300m due to south and then flies
- The radioactivity of a certain material drops to 1/16 of the initial value in 2 hours.
- For CE transistor amplifier, the audio signal voltage across the collector resistance
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply