The Bohr model of the hydrogen atom states that the single...

The Bohr model of the hydrogen atom states that the single electron can exist only in certain allowed orbits around the proton. The radius of each Bohr orbit is $r=n^{2}(0.0529 \mathrm{nm})$ where $n=1,2,3, \ldots .$ Calculate the electric potential energy of a hydrogen atom when the electron is in the (a) first allowed orbit, with $n=1 ;$ (b) second allowed orbit, with $n=2 ;$ and (c) when the electron has escaped from the atom, with $r=\infty$. Express your answers in electron volts.

Transcript

00:01 Hello students in this question we have to calculate the electric potential energy of a hydrogen atom for the given cases.

00:08 So we note that the electric potential energy up is equals to minus k e to the power 2 over r is the formula for the electric potential energy and here the value of r is given to us as n squared multiplied with 0 .0 529 nanometers so this is the value of r now in the first part which is a the value of n is equals to 1 so the electric potential energy will be equals to minus the value of k is 9 multiplied with 10 to the power 9 multiplied by the charge on an electron which is 1 .6 multiplied by the charge with 10 to the power minus 19 whole square over the value of r will be 1 square multiplied by 0 .0 529 multiplied by 10 to the power minus 9 to convert it into meters because it was nanometers and to convert it into electron volts we'll again divide it with 1 .6 multiplied with 10 to the power minus 19 so it will come out equals to minus 27 .22 electron volts now in the b part we are to calculate it for n equals to 2 so u p is equal to 9 multiplied with 10 to the power 9 multiplied with 1 .6 multiplied with 10 to the power minus 19 whole square over 2 square multiplied with 0 .0 529 multiplied with 10 to the power minus 9 and we will again divide it with 1 .6 multiplied with 10 to the power minus 19 to convert it into electron over so it comes out to echo to minus 6 .81 electron volts.

02:05 Now in the c part we have to calculate it for n equals to infinity so when the electron is escaped so it will be u p is equal to minus k e square over r can be written as n squared multiplied with 0 .0 5 to 9 nanometers so on putting n equals to infinity here the potential energy will come out equals to 0 so this is 0 in this case now let us write our other for all the parts, so our answer to the question is, for the part a is minus 27 .2 electron volts...

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