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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
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Topics: Dual Nature of Matter and Radiation
(150)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A rod of 10 cm length is moving perpendicular to uniform magntic field of intensity
- In an inductor when current changes from 2 A to 18 A in 0.05 sec, the e.m.f. induced
- When the energy of the incident radiation is increased by 20%,
- Huygen’s wave theory allows us to know
- The binding energy per nucleon of ₃⁷Li and ₂⁴He nuclei are 5.60 MeV and 7.06MeV
Topics: Dual Nature of Matter and Radiation (150)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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