Question

Quantum Mechanical Atom QUANTUM MECHANICAL ATOM Welcome to this IE. You may navigate to any page you've seen already using the IE Outline tab on the right. The orbital quantum number for the electron in a hydrogen atom is 1 = 6. What is the smallest possible value for the total energy of this electron? Emin = eV Submit

Answer 1

Answer:- $\inline \small \mathbf{E_{min} = - 0.2776 \; eV}$

Explanation:-

Orbital quantum number:-

              The orbital quantum number $\small l$ determines the shape of an orbital of an atom and the angular distribution. Each value of orbital quantum number $\small l$ indicates a specific s, p, d, f subshell. The value of orbital quantum number $\small l$ is dependent on the principal quantum number $\small n$. Orbital quantum number $\small l$ can be a positive integer, but it cannot be larger than one less than the principal quantum number $\inline \small \left ( n-1 \right )$ :

$\inline \small l = 0, 1, 2, 3, 4, . . . . , \left ( n-1 \right )$

In the given question value of orbital quantum number $\small l$ is 6, so principle quantum number will be,

$\inline \small As\; \; \; \; l =6$

$\inline \small l = \left ( n-1 \right )$

$\inline \small \therefore \; \; \; \; n=7$

Now we know the value of principle quantum number $\inline \small n=7$ . For the hydrogen atom and other nuclei having one electron, the total energy depends only upon the principal quantum number $\small n$.

Smallest possible value, also called ground state energy of electron, for the total energy of the electron in $\inline \small n^{th}$ orbit is given as,

$\small E_{min} = - \frac{Z^{2}\, e^{4}\, m_{e}}{8\, \epsilon _{0}\, h^{2}}\left ( \frac{1}{n^{2}} \right )$

Where,

• $\small Z$ is the atomic number of the atom and for hydrogen it is 1.
• $\small e$ is the charge of an electron in eV. $\small e= 1.6 \times 10^{-19} \: \: eV$
• $\small m_{e}$ is the mass of an electron in kg.   $\small m _{e}= 9.1 \times 10^{-31} \: \: kg$
• $\small \epsilon _{0}$ is the permittivity of the free space. $\small \epsilon _{0}=8.854\times 10^{-12} \: \: Farad/m$
• $\small h$ is the Planck's constant in Js.    $\small h=6.626\times 10^{-34} \: \: Js$
• $\small n$ is the principle quantum number. $\small n = 7$

So by plugging all the values in the above formula we get,

$\small E_{min} = - \frac{1^{2}\times \left (1.6\times10^{-19} \right )^{4}\times9.1\times10^{-31}}{8\times \left (8.854\times10^{-12} \right )^{2}\times \left ( 6.626\times10^{-34} \right )^{2}}\left ( \frac{1}{n^{2}} \right )$

$\small E_{min} = - \frac{6.554\times10^{-76}\times 9.1\times 10^{-31}}{8\times \left (78.4\times10^{-24} \right )\times \left ( 43.9\times10^{-68} \right )}\left ( \frac{1}{n^{2}} \right )$

$\small E_{min} = - \frac{59.64\times10^{-107}} {\left ( 27534.08\times10^{-92} \right )}\left ( \frac{1}{n^{2}} \right )$

$\small E_{min} = - 0.002166\times 10^{-15}\left (\frac{1}{n^{2}} \right )\: \: J$

    in by converting above value from joule to eV we get,

$\small E_{min} = \left (- 13.6 \right ) \frac{1}{n^{2}}$

$\small E_{min} = -13.6\times \frac{1}{49}$

$\small \mathbf{E_{min} = - 0.2776 \; eV}$

Therefore, for the electron in hydrogen atom $\inline \small l=6$ , the minimum possible value for total energy of the electron is -0.2776 eV.