2.51  The work function for caesium atom is 1.9 eV. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

Answers (1)
D Divya Prakash Singh

Finding the kinetic energy of the ejected electrons:

K.E of the ejected photoelectron:

= h[\nu-\nu_{o}] = hc\left [\frac{1}{\lambda} -\frac{1}{\lambda_{o}} \right ]

=(6.626\times10^{-34}Js)(3.0\times10^8m/s)\left [ \frac{1}{500\times10^{-9}m} - \frac{1}{654\times10^{-9}m} \right ]

= 19.878\times10^{-26}Jm\left [ \frac{154}{327000}\times10^9 m^{-1}\right ]

= 9.361\times10^{-20}J

Finding the Velocity of the ejected electrons:

KE = \frac{1}{2}mv^2

Where, 

m = mass of electron

v = velocity of electron

Therefore, the velocity is given by,

v =\sqrt{ \frac{2KE}{m}}

v = \sqrt{\frac{2\times9.361\times10^{-20}J}{9.1\times10^{-31}kg}} = 4.52\times10^5\ m/s

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