An element undergoes a reaction as shown `sx+2e^(-)tox^(-2)`
Energy released `=30.87` ev/atom. If the energy released is used to dissociated `4g` to `H_(2)` molecules equally into `H^(+)` and `H^(+)` is excited state of `H` atoms where the electron travels in orbit whose circumference equal to four times its de -roglie's wavelength. Determine the minimum number of moles of `x` that would be required.
Given IE of `H=13.6` ev/atom, bond energy of `H_(2)=4.526`v/molecule

An element undergoes s reaction as show - X+2e^(-) rarrX^(2-) , energy relesed =30.87eV//"atom" . If the energy released ,is used to dissociate 8 gms of H_(2) molecules,equally into H^(+) and H^(ast) ,where H^(ast) is an exicted state of H atoms where the electron travels in orbit whose circumference equal to four times its de Broglie's wavelenght . (a) Calculate the excited statue of H (b) Determine the least amount of X that would be required . Given : IE of H = 13.6 eV//"atom" , Bond energy of H_(2) =4.526 eV//"molecule" .

An element undergoes a reaction as shown : X+e^(-) to X^(-) Energy released =30.876 eV The energy released, is used to dissociate 8 g of H_2 molecules equality into H^+ and H^* , where H^* is in an excited state, in which the electron travels a path length equal to four times its debroglie wavelength. (a)Determine the least amount (moles) of 'X' that would be required. Given : I.E. of H =13.6eV/atom Bond energy of H_2 =4.526 eV/molecule.

Energy of electron in a orbit of H -atom is

The velocity of an e^- in excited state of H atom is 1.093xx10^6 m/s what is the circumference of this orbit?

The velocity of an electron in excited state of H-atom is 1.093 xx 10^6 m/s, what is the circumference of this orbit?

The velocity of an e in excited state of H-atom is 1.093 xx 10^(6)m//s , what is the circumference of this orbit?