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A radiation of wavelength 4.3 x 10^-7 m is shone on two different metal surface, nickel and potassium . (Speed of light 3.0 x10^8m/s, plank’s...

      

A radiation of wavelength 4.3 x 10^-7 m is shone on two different metal surface, nickel and potassium . (Speed of light 3.0 x10^8m/s, plank’s constant h as 6.63 x 10^-34 Js
i. Determine the energy of the incident radiation.
ii. If the work function of nickel is 8.0 x 10^-19J and that of potassium 3.68 x 10^-19J, which of the two metal will the given light eject electrons, state the reason.
iii. Calculate the velocity of the emitted electrons from the metal surface in (ii) above. (take the mass of an electron as 9.1 x 10^-31kg)

  

Answers


joseph
E=hc/wavelength
= (6.63 x 10^-34)(3.0 x 10^8)/(4.3 x 10^-7)
=4.626 x 10^-19 J
Potassium. The reason being the work function of potassium is lower than the energy of incident radiation.
E= w0 + K.E.
K.E. =E-W0
=4.63 x 10^-19 -3.68 x 10^-19
=9.46 x 10^-20 J
½ meV^2 =9.46 x 10^-20
V =v(2.087 X 10^11)
=4.569 x 10^5 m/s

joseph rimiru answered the question on December 10, 2017 at 18:27


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