According to Bohr's theory of hydrogen-atom, for the electron in the $n^{t h}$ permissible orbit

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Answer 1

Angular Momentum Quantisation:
$m v r = \frac{n h}{2 π}$ ------------(1)

Centripetal force:
$\frac{k ( e ) e}{r ^{2}} = \frac{m v ^{2}}{r}$ ----------------(2)
$v^{2} r = \frac{k e ^{2}}{m}$

Substituting (1) in (2)
$v = \frac{2 π k e ^{2}}{n h}$

Energy: $E = 13.6 \frac{Z ^{2}}{n ^{2}}$ -----------(3)

Linear Momentum:
$m v r = n h / 2 π$
$m v = \frac{n h}{2 π r}$ ---------------(4)

From the concept attached:
$r \propto n^{2}$ -----------------(5)

Hence, $m v \propto \frac{1}{n}$ by (6)
and
$U = 2 K E$
$E \propto \frac{1}{n ^{2}}$ (concept attached)

Answer 2

Angular Momentum Quantisation:

m v r = \frac{n h}{2 π}

------------(1)

Centripetal force:

\frac{k ( e ) e}{r ^{2}} = \frac{m v ^{2}}{r}

----------------(2)

v^{2} r = \frac{k e ^{2}}{m}

Substituting (1) in (2)

v = \frac{2 π k e ^{2}}{n h}

Energy:

E = 13.6 \frac{Z ^{2}}{n ^{2}}

-----------(3)

Linear Momentum:

m v r = n h / 2 π

m v = \frac{n h}{2 π r}

---------------(4)

From the concept attached:

r \propto n^{2}

-----------------(5)

Hence,

m v \propto \frac{1}{n}

by (6)
and

U = 2 K E

E \propto \frac{1}{n ^{2}}

(concept attached)

According to Bohr's theory of hydrogen-atom, for the electron in the $n^{t h}$ permissible orbit

$Linear Momentum \propto \frac{1}{n}$

$Radius of orbit \propto n$

$Kinetic Energy \propto \frac{1}{n ^{2}}$

$Angular Momentum \propto n$

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According to Bohr's theory of hydrogen-atom, for the electron in the $n^{t h}$ permissible orbit

$Linear Momentum \propto \frac{1}{n}$

$Radius of orbit \propto n$

$Kinetic Energy \propto \frac{1}{n ^{2}}$

$Angular Momentum \propto n$

In a hydrogen atom, electron is in the $n^{t h}$ excited state. It comes down to the first excited state by emitting $10$ different wavelengths. The value of $n^{t h}$ is

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