What wavelength of light is necessary to produce photoelectron of speed 2×108mswith a magnesium target?

Short Answer

Expert verified

The wavelength of light necessary to produce photoelectron of speed 2×108ms with a magnesium target is 82.5 nm.

Step by step solution

01

Converting into joules.

The energy of the photon is, E=hcλ . The Kinetic energy of the photon from the zinc plate is given by: KEmax=E-φ Here φ is the work function.

Now reforming the expression, we have,

λ=hc122+f

Because magnesium's work function potential is 3.7 eV, it must first be converted to Joules:

(3.7eV)1.6×1014 J1 cV=5.92×1011J

02

Determine the value of wavelength.

Substituting the value to calculate the value of wavelength,

λ=(6.63×1014 Js)3×102ms12(9.1×1013 kg)2×106ms2+5.92×1017 J=8.25×108 m=82.5nm

Hence the value of wavelength is 82.5 nm.

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