Structure of Atom

We know Rydberg formula,

${1 \over \lambda } = R \times {Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$

[ for hydrogen atom Z = 1 ]

     = $1.097 \times {10^7}\left( {{1 \over {{1^2}}} - {1 \over {{\infty ^2}}}} \right)$

$ \therefore $ $\lambda $ = ${1 \over {1.097 \times {{10}^7}}}$

= 9.11 $ \times $10^-8 m

= 91.1 $ \times $10^-9 m

= 91.1 nm

[ as 1 nm = 10^-9 m ]

Isoelectronic means those species whose no of electrons are same.

Ions	Atomic No	No of Electron
K+	19	19-1=18
Ca+2	20	20-2=18
Sc+3	21	21-3=18
Cl-	17	17+1=18

So, option $(C)$ is correct.

Isoelectronic means those species whose electron number are same.

Species	Atomic Number	Electron Number
N3-	7	7 + 3 = 10
F-	9	9 + 1 = 10
Na+	11	11 - 1 = 10
Be	4	4
Al3+	13	13 - 3 = 10
Cl-	17	17 + 1 = 8
Ca2+	20	20 - 2 = 18
Cs+	55	55 - 1 = 54
Br	35	35
Mg2+	12	12 -2 = 10

$\therefore$ N^3-, F^-, Na⁺

As per de-Broglie wavelength of a particle is

Wavelength ($\lambda $) = ${h \over {mv}}$

Where h is plank's constant, m is mass of the particle and v is the velocity of the particle

$\therefore \lambda $ = ${{6.63 \times {{10}^{ - 34}}} \over {0.06 \times 10}}$

= ${{6.63 \times {{10}^{ - 34}} \times 10} \over {0.6 \times 10}}$

= ${{6.63 \times {{10}^{ - 33}}} \over 6}$

$ \cong 10{}^{ - 33}$

In Bohr series of lines of hydrogen spectrum :- Red end means end of visible region which is balmer series of lines so n will be 2

Now for third lines with respect to Balmer series means n = (2 + 3) = 5^th line

$\therefore$ n₁ = 2 and n₂ = 5

For 2 $\to$ 5 electron will need energy to jump as 2 > 5 so no spectrum will release instead energy will be absorbed.

$\therefore$ 5 $\to$ 2

For s-electron l = 0

$\therefore $ $\sqrt {l(l + 1)} {h \over {2\pi }}$

= $\sqrt {0(0 + 1)} {h \over {2\pi }}$

= $\sqrt 0 {h \over {2\pi }}$ = 0 (zero)

According to Hysenberg's Uncertainty Principal $$\Delta x . \Delta p \ge { h \over {4\pi}}$$ Where $\Delta x $ is uncertainty in position and $\Delta p $ is uncertainty in momentum and as in momentum m (mass of object) is always same so we can say $$\Delta x . m. \Delta v \ge { h \over {4\pi}}$$ Mass of the particle (m) = 25 g = 0.025 kg

Position of the particle ($\Delta x$) = 10^-5 m

$\Delta x . \Delta p $ = ${ h \over {4\pi}}$

$\Delta x . m . \Delta v $ = ${ h \over {4\pi}}$

$\Delta v$ = ${h \over {4 \times \Delta x \times m \times \pi }}$

$\Delta v$ = ${{6.6 \times {{10}^{ - 34}}} \over {4 \times {{10}^{ - 5}} \times 0.025 \times 3.14}}$

$\Delta v$ = 2.1 $\times$ 10^-28

n = 1 means ground state

n = 2 means 1^st excited state

n = 3 means 2^nd excited state

Remember: n^th excited state means n = n + 1

We know energy of an atom is given by

Energy (E) = -13.6 $\times$ ${{{Z^2}} \over {{n^2}}}$

= -13.6 $\times$ $1 \over 9$ [ $\because$ Z = atomic number = 1 for hydrogen atom]

= -1.51 ev