| ⇦ | 
| ⇨ |
An electron in the hydrogen atom jumps from excited state n to the ground state. The wavelength so emitted illuminates a photosensitive material having work function 2.75 eV. If the stopping potential of the photoelectron is 10 V, the value of n is
Options
(a) 3
(b) 4
(c) 5
(d) 2
Correct Answer:
4
Explanation:
KEₘₐₓ = 10 eV ? = 2.75 eV
Total incident energy
E = ? + KEₘₐₓ = 12.75 eV
Energy is released when electron jumps from the excited state n to the ground state.
E₄ – E₁ = {-0.85 – (-13.6) eV}
= 12.75 eV value of n = 4.
Related Questions: - The numebr 0 (zero) is required for
- The sodium nucleus ₁₁Na²³ contains
- If the kinetic energy of the particle is increased to 16 times its previous value
- The horizontal range and maximum height of a projectile are equal.The angle of projection
- The mean free path of electrons in a metal is 4 x 10⁻⁸ m. The electric field
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The numebr 0 (zero) is required for
- The sodium nucleus ₁₁Na²³ contains
- If the kinetic energy of the particle is increased to 16 times its previous value
- The horizontal range and maximum height of a projectile are equal.The angle of projection
- The mean free path of electrons in a metal is 4 x 10⁻⁸ m. The electric field
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