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  • If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5 x 10^7 m s -1, calculate the energy with which it is bound to the nucleus.

2.54: If the photon of the wavelength 150 pm strikes an atom and one of it's inner bound electrons is ejected out with a velocity of 1.5\times 10^7\ \textup{m s}^{-1} , calculate the energy with which it is bound to the nucleus.

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Given the wavelength of a photon which strikes an atom is 150\ pm = 150\times10^{-12} m.

Then the energy associated with this photon will be:

E= \frac{hc}{\lambda} = \frac{(6.626\times10^{-34}Js)(3.0\times10^8)m/s}{150\times10^{-12}m}

= 1.3252\times10^{-15}J

Given the velocity of ejected inner bounded electron: v= 1.5\times10^7 m/s .

Then, the energy associated with this electron will be, Kinetic energy.

Hence finding

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

Where 

m = mass of electron, v = velocity of electron

KE = \frac{1}{2}(9.11\times10^{-31}kg)(1.5\times10^7)^2

= 1.02\times10^{-16}J

Hence the energy with which the electrons are bounded to the nucleus is:

= 13.25\times10^{-16}J - 1.02\times10^{16}J

= 12.23\times10^{-16}J

=\frac{ 12.23\times10^{-16}J}{1.602\times10^{-19}}

= 7.63\times10^3\ eV

 

 

 

 

 

Posted by

Divya Prakash Singh

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