Question

#Quality Question
@Atoms

Q) Derive Formulas of Radius,Speed,Time,Kinetic Energy,Potential Energy and Total Energy of Hydrogen like atom using Bohr model.

(Also write approx values of Radius, Speed and Time) ​

Answer 1

$\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese ANSWER}}}}$

$\mathfrak{Consider \: \: an \: \: electron \: \: revolving } \\ \mathfrak{ in \: \: circular \: \: orbit \: \: of \: \: radius \: \: \bold r } \\ \mathfrak{with \: \: Speed \: \: \bold v }$

$\mathcal{According \: \: To } \\ \large \maltese \: \: \underline \mathfrak{Bohr's \: \: 1st \: \: Postulate }$

$\underline\mathcal{ \bigstar \: \: F_{centripetal}= F_{electrosatic}} \\ \\ \bf \frac{mv {}^{2} }{r} = \frac{1}{4\pi \epsilon \circ} \frac{(ze)(e)}{ {r}^{2} } \\ \\ \boxed{ \bf {} m {v}^{2} = \frac{1}{4\pi \epsilon \circ} \frac{z {e}^{2} }{r} } \: \: \: \: \: ...(i)$

$\mathcal{According \: \: To } \\ \large \maltese \: \: \underline \mathfrak{Bohr's \: \: 2nd \: \: Postulate }$

$\bf L = \frac{nh}{2\pi} \\ \\ \sf mvr = \frac{nh}{2\pi} \\ \\ \boxed{ \bf v = \frac{nh}{2\pi mr} } \: \: \: \: \: ...(ii)$

$\mathcal{Putting \: \: (ii) \: \: in \: \: (i)}$

$\sf m( \frac{nh}{2\pi mr} ) {}^{2} = \frac{1}{4\pi \epsilon \circ} \frac{ {ze}^{2} }{r} \\ \\ \pink{ \boxed{ \boxed{ \mathfrak{r = \frac{ {n}^{2} {h}^{2} \epsilon \circ }{\pi mz {e}^{2} } }}}} \: \: \: \: \: ...(iii)$

$\mathcal{Putting \: \: (iii) \: \: in \: \: (ii)}$

$\pink{ \boxed{ \boxed{ \mathfrak{v= \frac{ {ze}^{2} }{2nh \epsilon \circ} }}}}$

$\huge\mathcal{As}$

$\bf T = \frac{2\pi r}{v} \\ \\ \bf \qquad \qquad\mathfrak{putting \: \: both \: values} \\ \\ \pink{ \boxed{ \boxed{ \mathfrak{t = \frac{4 {n}^{3} {h}^{3} { \epsilon \circ}^{2} }{m {z}^{2} {e}^{4} } }}}}$

$\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: \: \: FROM (i)}}}}$

$\bf m {v}^{2} = \frac{1}{4\pi \epsilon \circ} \frac{ {ze}^{2} }{r}$

$\pink{ \boxed{ \boxed{ \mathfrak{ Kinetic Energy = \frac{1}{8\pi \epsilon \circ} \frac{ {ze}^{2} }{r} }}}}$

$\bf Potential \: Energy = \frac{1}{4\pi \epsilon \circ } \frac{(ze)( - e)}{r} \\ \\ \pink{ \boxed{ \boxed{ \mathfrak{ Potential \: \: Energy = - \frac{1}{4\pi \epsilon \circ} \frac{ {ze}^{2} }{r} }}}}$

$\bf Total \: Energy = Kinetic \: Energy \: + Potential \: Energy \\ \\ = \sf Kinetic \: Energy - 2( Kinetic \: Energy ) \\ \\ = \sf - (Kinetic \: Energy )$

$\pink{ \boxed{ \boxed{ \mathfrak{ Total \: Energy = - \frac{1}{8\pi \epsilon \circ} \frac{ {ze}^{2} }{r} }}}}$

$\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: \: Approx \: \: Values }}}}$

$\bf \: r = 0.529 \times \frac{ {n}^{2} }{z}A° \\ \\ \bf v = 2.18 \times {10}^{6} \times \frac{z}{n} m/s \\ \\ \bf T = 1.53 \times 10 {}^{ - 10} \times \frac{{n}^{3} }{ {z}^{2} }second$

Answer 2

Answer:

See Attachment And Get your Answer