HSC Science (General) 12th Board ExamMaharashtra State Board
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Find the Maximum Kinetic Energy of the Photoelectrons in Joules. - HSC Science (General) 12th Board Exam - Physics

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Question

The photoelectric current in a photoelectric cell can be reduced to zero by a stopping potential of 1.8 volt. Monochromatic light of wavelength 2200Å is incident on the cathode. Find the maximum kinetic energy of the photoelectrons in joules. [Charge on electron = 1.6 x 10-19 C]

Solution

Given:- V0 = 1.8 V, e = 1.6 x 10-19 C, λ = 2200 Å

To find:- Maximim kinetic energy (K.E.)max

Formula:- (K.E.)max = eV0

Calculation: Using formula,

(K.E.)max = 1.6 x 10-19 x 1.8

∴ (K.E)max = 2.88 x 10-19 J

Maximum kinetic energy of emitted photoelectron is 2.88 x 10-19 J.

  Is there an error in this question or solution?

APPEARS IN

 2014-2015 (October) (with solutions)
Question 5.2 | 7.00 marks
Solution Find the Maximum Kinetic Energy of the Photoelectrons in Joules. Concept: Photoelectric Effect - Hertz’S Observations.
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