question_answer 73)

If photon of wavelength 150 pm strikes on an atom and one of its inner bound electrons is ejected out with velocity of\[1.5\times {{10}^{7}}m\,{{\sec }^{-1}}\]. Calculate energy with which it is bound to the nucleus.

Answer:

Energy of photon \[=\frac{hc}{\lambda }\frac{6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}}{150\times {{10}^{-12}}}\] \[=1.325\times {{10}^{-15}}\text{J}\] Kinetic energy of ejected electron \[=\frac{1}{2}m{{v}^{2}}\] \[=\frac{1}{2}\times 9.1\times {{10}^{-31}}\times {{(1.5\times {{10}^{7}})}^{2}}\] \[=1.0237\times {{10}^{-16}}\text{J}\] Binding energy of electron to the nucleus \[=1.325\times {{10}^{-15}}-1.0237\times {{10}^{-16}}\] \[=1.2226\times {{10}^{-15}}\text{J}\] \[=\frac{1.2226\times {{10}^{-15}}}{1.6\times {{10}^{-19}}}eV\] \[=7.64\times {{10}^{3}}eV\]