| ⇦ | 
| ⇨ |
The threshold frequency for a photosensitive metal is 3.3 x 10¹⁴ Hz. If light of frequency 8.2 x 10¹⁴ Hz incident on this metal, the cut-off voltage for the photoelectric emission is nearly
Options
(a) 2 V
(b) 3 V
(c) 5 V
(d) 1 V
Correct Answer:
2 V
Explanation:
K.E = hv – hvₜₕ = eV₀ ( V₀ = cut off voltage)
V₀ = h/e (8.2 x 10¹⁴ – 3.3 x 10¹⁴)
= 6.6 x 10⁻³⁴ x 4.9 x 10¹⁴ / 1.6 x 10⁻¹⁹ = 2 V
Related Questions: - Which gate can be obtained by shorting both the input terminals of a NOR gate?
- An inductor is connected to an a.c. source when compared to voltage,
- A thin and circular disc of mass M and radius R is rotating in a horizontal plane
- The charge flowing through a resistance R varies with time t as Q=at-bt²,
- If threshold wevelength for sodium is 6800 Å, the work function will be
Topics: Dual Nature of Matter and Radiation
(150)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Which gate can be obtained by shorting both the input terminals of a NOR gate?
- An inductor is connected to an a.c. source when compared to voltage,
- A thin and circular disc of mass M and radius R is rotating in a horizontal plane
- The charge flowing through a resistance R varies with time t as Q=at-bt²,
- If threshold wevelength for sodium is 6800 Å, the work function will be
Topics: Dual Nature of Matter and Radiation (150)
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