Question

Question (2) Choose the correct answer. Justify your answer In a photoelectric experiment, the stopping potential is 1 V, the work function is 2 eV. The intensity of the incident light is doubled and the frequency reduces to half its initial value. Answer the following two questions 1) In this new situation, the work function of the cathode is: a) 4 eV b) 2 eV l eV d) 6 eV e) 3 eV

Answer 1

Work function in the property of the metal and in independent of intensity or frequency of light So, Answer-B > 2ev saturation current depends only on intensity of light and not the frequency of light when intensity of incident light double, number of electron hitting the metal target or cathode doubles. So, saturation current also double From Compton scattering, the scattered wavelength in greater than incident wave length \r\n delta lambda = h/m_ec (1 - cos emptyset) lambda_1 - lambda_2 = h/m_ec (1 - cos emptyset) (2.21 - 1)_pm = 6.626 times 10^-34/9.1 times 10^-31 times 3 times 10^8 (1 - cos emptyset) = 0.243 times 10^-11 1 - cos emptyset = 1.21 times 10^-12/0.243 times 10^-11 = 0.5 cos emptyset = 0.5, emptyset = 60 degree K.E = 20 m_0 c^2 = (squareroot - 1) m_0 c^2 squareroot - 1 = 20: r = 21 = 1/ squareroot 1 - v^2/c^2 1 - v^2/c^2 = 1/441 rightdoublearrow v^2/c^2 = 1 - 1/441 = 440/441 v = 20.976/21 c = 0.9989 c v = 0.998 c v = v_0 (1 +
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