Question

For the first orbit of hydrogen atom the minimum excitation potential is:
$\ A. \ 13.6 \ V$
$\ B. \ 3.4 \ V$
$\ C. \ 10.2 \ V$
$\ D. \ 3.6 \ V$

Answer 1

Minimum excitation potential corresponds to the minimum potential energy. Minimum potential energy change occurs when an electron jumps from $\ n=1$ to $\ n=2$. For any other combination, energy won’t be minimum. Also, excitation always means going to higher energy levels. That’s why we concluded that the jump is from $\ n=1$ to $\ n=2$.

Formula used:
${ E }_{ n }=\dfrac { 2\pi ^{ 2 }mK^{ 2 }Z^{ 2 }e^{ 4 } }{ n^{ 2 }h^{ 2 } }$
Where, $\ m$ = mass of electron = $\ 9.1 \times 10^{-31} kg$
$\ K$ =$\dfrac { 1 }{ 4\pi { \varepsilon }_{ o } }$= $\ 8.987 \times 10^9 kgm^3s^{-2}C^{-2}$
$\ Z$ = Atomic number of the element (for hydrogen$z=1$)
$\ e$ = Charge on electron =$\ -1.602 \times 10^{-19}$ C
$\ h$ = Planck's constant = $\ 6.626 \times 10^{-34} m^2 kg / s$
$\ n$ = orbit number
& Electric potential ( $\ V$) = Potential energy $\ ( E )$ $\times$Charge ($\ e$)

Complete step by step answer:
In order to find the value of energy needed for a transition to happen, let’s use the formula,
${ E }_{ n }=\dfrac { 2\pi ^{ 2 }mK^{ 2 }Z^{ 2 }e^{ 4 } }{ n^{ 2 }h^{ 2 } }$ and put all the known values:
${ E }_{ n }=\dfrac { 2\pi ^{ 2 }\times 9.1 \times 10^{-31} \times({8.987 \times 10^9})^2 1^{ 2 }\times ({1.602 \times 10^{-19}} )^{ 4 } }{ n^{ 2 }\times ({6.626 \times 10^{-34}})^{ 2 } }$= $\dfrac { 13.6 }{ n^{ 2 } }$ $\ eV$
For minimum potential energy difference, electrons must just go from $\ n=1$to $\ n=2$ during the excitation process.
Hence, $\ {E}_{1} \quad - \quad {E}_{2}$= $\dfrac { 13.6 }{ 1^{ 2 } } - \dfrac { 13.6 }{ 2^{ 2 } }$= $\ 13.6 - 3.4 = 10.2$ $\ eV$
Hence, minimum potential = $\dfrac{E}{e} = 10.2 V$

Hence the correct option is C.

Additional Information:
For modern physics, it is suggested to learn formulas for
a> Bohr’s radius – Which means the radius in which electrons revolve in an atom.
b> Velocity in $\ n^{th}$ orbit – Which means the velocity of electrons in $\ n^{th}$ orbit
c> Energy of $\ n^{th}$ orbit – Which means the energy of the electron in $\ n^{th}$ orbit . Deriving them in the exam could consume time of students during examination.

Note:
It’s not possible to calculate such complicated results of energy. Hence students are advised to remember formula $\ E_{n}$=$\dfrac { 13.6 }{ n^{ 2 } }$ $\ eV$ . Along with it, the values corresponding to $\ n=1$, $\ n=2$ i.e. 13.6$\ eV$, 3.6$\ eV$, etc. are frequently used in modern physics. Remembering these can save a lot of time in examination.
Possibility of mistake is that students might consider minimum potential energy for$\ n=1$, but here we have to see the minimum energy change during transition (excitation), which is $\ n=1$ to$\ n=2$.