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Manipal MET2022PhysicsDual Nature of Matter

The electric field of light wave is given is E=10^3 ( 2 x 5 10⁻⁷ -2 6 10¹⁴ t ) j N / C . This light falls on a metal plate of work function 1.5 eV . The stopping potential of the photoelectron is V . (Energy of photon = 1240 ( in nm ) eV ).

Correct answer

0.98

Step-by-step solution

E=E₀ (k x- t) =2 6 10¹⁴ array ll & f= 2 = 2 6 10¹⁴ 2 =6 10¹⁴ ~Hz & = Energy of photon = 1240 ( in nm ) eV & = c f = 3 10^8 6 10¹⁴ =0.5 10⁻⁶ ~m =500 ~nm & = 1240 500 eV =2.48 eV array Now, by Einstein photoelectric equation, array rlrl & & & = ₀+ KE = ₀+e V_s ( KE =e V_s ) & 2.48 & =1.5+e V_s & & e V_s & =2.48-1.5=0.98 eV & V_s & =0.98 ~V array

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