Question 54

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.

Answer

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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