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 $$1.5 \times 10^{7}\, ms^{-1}$$, calculate the energy with which it is bound to the nucleus.

Asked by Abhisek | 1 year ago |  66

##### Solution :-

Energy of incident photon (E) is given by,

$$E=\dfrac{hc}{\lambda }\\ \\ E=\dfrac{(6.626\times 10^{-34})(3\times 10^{8})}{(150\times 10^{-12})}$$

$$=1.3252\times 10^{-15}\, J \\ \\ \simeq 13.252\times 10^{-16}J$$

Energy of the electron ejected (K.E)

$$\dfrac{1}{2}m_{e}\nu ^{2}\\ \\ =\dfrac{1}{2}(9.10939\times 10^{-31}kg)(1.5\times 10^{7}ms^{-1})^{2}\\ \\ =10.2480\times 10^{-17}J\\ \\ =1.025\times 10^{-16}J$$

Therefore , the energy that binds the electron to the nucleus can be determined using the following formula:

= E – K.E

$$=13.252\times 10^{-16}J-1.025\times 10^{-16}J\\ \\ =12.227\times 10^{-16}J=\dfrac{12.227\times 10^{-16}}{1.602\times 10^{-19}}eV\\ \\ =7.6\times 10^{3}eV$$

Answered by Pragya Singh | 1 year ago

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