# A clean nickel surface (work function 5.1eV), is exposed to light of wavelength 206nm. What is the maximum speed of the photoelectrons emitted from this surface?

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vrma Share
Mar 9, 2018

$\frac{1}{2} M . V {m}^{2} = h . \frac{c}{\lambda} - \Phi$
The Einstein relation can be used.

#### Explanation:

The incident light on nickel surface can provide photon energy

to an electron of the metal which is equal to planck's constant
times the frequency of light.

here we have wavelength given so frequency= velocity of light/wavelength.

One can use the photoelectric equation given above to calculate maximum speed of photoelectrons

Under the assumption that the whole photon energy is used up

First -in taking out electron from the metal (work function)

Secondly -the left over is provided as the kinetic energy to the photo-electron

When one puts in the numbers ,the maximum speed can be of the order of ...10^5 meters /s

**$V \left(m\right) =$ sqrt { 2/M ( hc/ lambda)- 5.1 eV }

where M= mass of electron = 9.11 x 10^-31 kg.
h= Planck's Constant =6.63x 10^-34 Js
c= speed of light = 3x 10^ 8 m/s
lambda= wavelength of light = 206 nm**

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