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  • 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.

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)

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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

Posted by

Divya Prakash Singh

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