If the photon of the wavelength 150 nm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5 X 10⁷ ms^-1 calculate the energy with which it is bound to the nucleus.

Solution

space space Here space straight lambda space equals space 150 space pm space equals space 150 space cross times space 10 to the power of negative 12 end exponent straight m space Energy space of space photon space space space equals space hc over straight lambda space equals space fraction numerator left parenthesis 6.626 space cross times space 10 to the power of negative 34 end exponent thin space Js right parenthesis space cross times space left parenthesis 3 space cross times 10 to the power of 8 ms to the power of negative 1 end exponent right parenthesis over denominator 150 space cross times space 10 to the power of negative 12 end exponent straight m end fraction space space space space space space space space space space space space space space equals space 1.33 space cross times space 10 to the power of negative 15 end exponent straight J space Kinetic space energy space of space the space ejected space electron space equals space 1 half mv squared space space space space space equals space 1 half cross times 9.11 space cross times space 10 to the power of negative 31 end exponent space cross times space left parenthesis 1.5 space cross times space 10 to the power of 7 right parenthesis squared space space space space equals 1.025 space cross times space 10 to the power of negative 16 end exponent straight J

Binding energy = Energy of incident photon - kinetic energy

space space space space space space space equals space 1.33 space cross times 10 to the power of negative 15 end exponent straight J space minus 1.025 space cross times space 10 to the power of negative 16 end exponent straight J space space space space space space equals space 1.23 space cross times space 10 to the power of negative 15 end exponent straight J

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