Electromagnetic radiation of wavelength 285 nm is just sufficient to ionise the potassium atom. What is the ionization energy of potassium (in kJ mol^-1) ?

Solution

We know that energy of radiation (here it is the ionisation energy of potassium) is given by

straight E space equals space hv space equals space hc over straight lambda

...(1)

But space straight lambda space equals space 285 space nm space equals space 285 space cross times space 10 to the power of negative 9 end exponent straight m semicolon space space space space space space straight c space equals space 3.0 space cross times space 10 to the power of 8 space ms to the power of negative 1 end exponent space space space space space space straight h space equals space 6.626 space cross times space 10 to the power of negative 34 end exponent Js space space space space space space straight E space equals space fraction numerator 6.626 space cross times space 10 to the power of negative 34 end exponent Js space cross times 3.0 space cross times space 10 to the power of 8 space ms to the power of negative 1 end exponent over denominator 285 cross times 10 to the power of negative 9 end exponent straight m end fraction space space space space space space space space space space space equals 6.974 space cross times space 10 to the power of negative 19 end exponent straight J space per space atom or space space space straight E space equals space fraction numerator 6.974 space cross times space 10 to the power of negative 19 end exponent cross times space 6.02 space cross times space 10 to the power of 23 over denominator 10 cubed end fraction kJ space mol to the power of negative 1 end exponent space space space space space space space equals 420 space kJ space mol to the power of negative 1 end exponent.

Hence, ionisation energy of potassium is 420 kJ mol^-1

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