Electromagnetic radiation of wavelength 663nm is just sufficient to ionize the atom of metal A. The ionization energy of metal A in kJ mol-1 is…………… (Rounded-off to the nearest integer)
[h = 6.63 * 10-34Js,c = 3.00 * 108ms-1, NA = 6.02 * 1023 mol-1]
Electromagnetic radiation of wavelength 663nm is just sufficient to ionize the atom of metal A. The ionization energy of metal A in kJ mol-1 is…………… (Rounded-off to the nearest integer)
[h = 6.63 * 10-34Js,c = 3.00 * 108ms-1, NA = 6.02 * 1023 mol-1]
= 0.03 * 10-17 = 3.0 * 10-19 J/atom
= 18.06 * 101 kJ/mole = 180.6
Similar Questions for you
CH3COOH + NaOH → CH3COONa + H2O
ΔH = –50.6 kJ/mol
NaOH + SA [HCl] → NaCl + H2O
ΔH = –55.9 kJ/mol
the value of ΔH for ionisation of CH3COOH
⇒ ΔH = +55.9 – 50.6
5.3 kJ/mol
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