Q- 2.54: 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,
Energy of the electron ejected (K.E)
= 10.2480 × 10-17 J
= 1.025 × 10-16 J
Hence, the energy with which the electron is bound to the nucleus can be obtained as: = E - K.E
= 13.252 × 10-16 J - 1.025 × 10-16 J
Answer
Energy of incident photon (E) is given by,
Energy of the electron ejected (K.E)
= 10.2480 × 10-17 J
= 1.025 × 10-16 J
Hence, the energy with which the electron is bound to the nucleus can be obtained as: = E - K.E
= 13.252 × 10-16 J - 1.025 × 10-16 J
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