Question

Calculate the energy in kcal/mole necessary to remove an electron in a hydrogen atom in fourth principal quantum number to infinity.

Answer 1

To solve the given type of question, i.e. those questions which are only provided with the information about different states in which the electrons are present and to which the electrons get excited is solved using Rydberg's equation.
The equation is: $\dfrac{1}{\lambda }={{R}_{H}}\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right]$

Complete step-by-step answer: In the question, it is asked how much energy is required in kcal/mol to remove an electron present in a hydrogen atom in the fourth principal quantum number to infinity.
We are very familiar that the principal quantum represents the shell number and gives the idea in which shell the electron is present and it also gives an idea about the energy of the atom.
So here the fourth principal quantum number refers to the fourth shell$\left( n=4 \right)$, the electron is now present in the fourth shell of hydrogen and we have to remove the electron from the shell to infinity.
We can solve the problem using Rydberg's equation and the equation is as follows:
$\dfrac{1}{\lambda }={{R}_{H}}\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right]$
Here in the equation ${{n}_{1}}$and ${{n}_{2}}$represents two excited states of the atom. And the lambda ($\lambda$) refers to the wavelength associated with the excitations and ${{R}_{H}}$refers to Rydberg's constant.
So in this question,${{n}_{1}}=4$ and ${{n}_{2}}=\alpha$,${{R}_{H}}=10967800$
Substitute all the values in the equation and we will get the $\lambda$value.
$\dfrac{1}{\lambda }={{R}_{H}}\left[ \dfrac{1}{{{4}^{2}}}-\dfrac{1}{{{\alpha }^{2}}} \right]$
$\Rightarrow \dfrac{1}{\lambda }=\dfrac{{{R}_{H}}}{{{4}^{2}}}$
$\Rightarrow \dfrac{1}{\lambda }=\dfrac{{{R}_{H}}}{16}$
$\Rightarrow \lambda =\dfrac{16}{R}$
$\Rightarrow \lambda =\dfrac{16}{10967800}=14.59\times {{10}^{-7}}m$
We got the value of $\lambda$, now we have to calculate energy value for that we use the equation,
$E=\dfrac{hc}{\lambda }$
Where E refers to the energy, h refers to Planck's constant, c is the speed of light.
$h=6.626\times {{10}^{-34}}Js$ and $c=3\times {{10}^{8}}m/s$
Substitute the values in the equation:$E=\dfrac{6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}}{14.59\times {{10}^{-7}}}=1.36\times {{10}^{-19}}J$
We get the energy values in joules. We have to calculate this in Kilocalories so we have to do the conversions. And the value should be found for one mole of the substance.
Hence the equation becomes,
$E=1.36\times {{10}^{-19}}\times \dfrac{1}{4.18}\times {{10}^{-3}}\times 6.022\times {{10}^{23}}$
$\therefore E=19.634\dfrac{kcal}{mol}$

Note: We should be very thorough about the unit conversions to do such calculations. Here to convert Joules into kilocalorie we have used the conversions:
To convert joules to calories, divide the joule value by 4.18.
$4.18cal=1Joule$
To convert calories to the kilocalorie, divide the value by 1000.