HSC Science (General) 12th Board ExamMaharashtra State Board
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# Calculate the Maximum Kinetic Energy in Ev of Photoelectrons Emitted, When Ultraviolet Light of Wavelength 2600a Falls on It. - HSC Science (General) 12th Board Exam - Physics

ConceptEinstein’s Equation - Particle Nature of Light

#### Question

The threshold wavelength of silver is 3800Å. Calculate the maximum kinetic energy in eV of photoelectrons emitted, when ultraviolet light of wavelength 2600Å falls on it.

(Planck’s constant, h =6.63 x 1O-34J.s.,

Velocity of light in air, c = 3 x 108 m / s)

#### Solution

Wavelength of silver, λ1 = 3800 Aº

Wavelength of ultraviolet light, λ2 = 2600 Aº

H = 6.63 × 10-34 Js

Velocity of light in air, c = 3 × 108 m/s

To calculate kinetic energy, K.E. = hv − hvº

"K.E."="hc"[1/lambda-1/lambda_o]

"K.E."=19.89xx10^-19[1/2.6-1/3.8]

"K.E."=(2.416xx10^-19)/(1.6xx10^-19)

K.E. = 1.51 eV

Hence, the maximum kinetic energy emitted by the photoelectron is 1.51 eV.

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#### APPEARS IN

2014-2015 (March) (with solutions)
Question 7.4 | 3.00 marks

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Solution Calculate the Maximum Kinetic Energy in Ev of Photoelectrons Emitted, When Ultraviolet Light of Wavelength 2600a Falls on It. Concept: Einstein’s Equation - Particle Nature of Light.
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