The electric field associated with a monochromatic beam becomes zero 1.2 × 10^15 times per second. Find the maximum kinetic energy of the photoelectrons when this light falls on a metal surface whose work function is 2.0 eV.

Question

Deepak Patra · Answer

Sol. The electric field becomes 0 1.2 * 10^45 times per second. ∴ Frequency = 1.2 * 10^15/2 = 0.6 * 10^15 hv = ∅ base 0 + KE ⇒ hv - ∅ base 0 = KE ⇒ KE = 6.63 * 106-34 *0.6 * 10^15/1.6 * 10^-19 – 2 = 0.482 ev = 0.48 ev.

Akanksha · Answer

In one complete vibration twice the electric field becomes zero so the frequency of incident light is given by Frequency = 1/2*2.4*10^15 HzMaximum K. E. = h*frequency- work function 6.63*10^-34*1.2*10^15/1.6*10^-19=2.97eV

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The electric field associated with a monochromatic beam becomes zero 1.2 × 10^15 times per second. Find the maximum kinetic energy of the photoelectrons when this light falls on a metal surface whose work function is 2.0 eV.

The electric field associated with a monochromatic beam becomes zero 1.2 × 10^15 times per second. Find the maximum kinetic energy of the photoelectrons when this light falls on a metal surface whose work function is 2.0 eV.

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The electric field associated with a monochromatic beam becomes zero 1.2 × 10^15 times per second. Find the maximum kinetic energy of the photoelectrons when this light falls on a metal surface whose work function is 2.0 eV.

The electric field associated with a monochromatic beam becomes zero 1.2 × 10^15 times per second. Find the maximum kinetic energy of the photoelectrons when this light falls on a metal surface whose work function is 2.0 eV.

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