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 × 107 ms–1, calculate the energy with which it is bound to the nucleus.
Energy of incident photon (E) is given by,
= 10.2480 × 10–17 J
= 1.025 × 10–16 J
Energy with which the electron was bound to the nucleus
= 13.25 × 10–16 J – 1.025 × 10–16 J
= 12.225 × 10–16 J
= 12.225 × 10–16 J/1.602 x10-19 eV
= 7.63x103eV
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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 54: If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is eject....
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