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Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. Ratio of maximum speed of emitted electrons will be
Options
(a) 1 : 4
(b) 1 : 2
(c) 1 : 1
(d) 1 : 5
Correct Answer:
1 : 2
Explanation:
The maximum kinetic energy of emitted electrons is given by
K.E = WorkFunction(₀) – WorkFunction(₁)
K.E₁ = 1 eV – 0.5 eV = 0.5 eV
K.E₂ = 2.5 eV – 0.5 eV = 2.0 eV
K.E₁ / K.E₂ = 0.5 eV / 2 eV = 1/4
KE = mv² /2
v₁ / v₂ = √1/4 = 1/2
Related Questions: - Photons of energy 6 eV are incident on a metal surface whose work function is 4 eV.
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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
- Photons of energy 6 eV are incident on a metal surface whose work function is 4 eV.
- The ratio of the acceleration for a solid sphere(mass m and radius R)rolling down
- A step-down transformer has 50 turns on secondary and 1000 turns on primary winding.
- In the given (V – T) diagram, what is the relation between pressure P₁ and P₂?
- The component of vector A=2i+3j along the vector i+j is
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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