Question

Derive an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n - 1). Also show that for
large values of n, this frequency equals to classical frequency of revolution
of an electron.

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Answer 1

$\large{\boxed{\underbrace{\sf{Question}}}}$

Derive an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n - 1). Also show that for large values of n, this frequency equals to classical frequency of revolution of an electron.

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$E_{1} = hv_{1} = \dfrac{hme_{4}}{(4\pi \: )^{3}€ ^{2}_{0}( \dfrac{h}{2\pi})^{ 3}} \times ( \dfrac{1}{n^{2}})$ Equation 1

Where,

• $v_{1}$ means Frequency of radiation at level n.

• h means Plank's constant.

• $€_{0}$ means Permitivily of the free space.

• m means Hydrogen atom's mass.

• e means charge of an electron.

Now, the relation of Energy $_{2}$ of the radiation at the level (n-1) is given as the following.

$E_{2} = hv_{2} = \dfrac{hme ^{4}}{(4\pi \:)^{3}€^{2} _{0}( \dfrac{h}{2\pi})^{3}} \times \dfrac{1}{(n - 1)}$ Equation 2

Where,

• $v_{2}$ means Radiation frequency at level (n-1)

Now, energy released as a result of de-excitation :

E = $E_{2}$ - $E_{1}$ hv = $E_{2}$ - $E_{1}$ Equation 3

Where,

• v means frequency of radiation emitted.

Now putting the values from Equation 1 and 2 in 3 we get the following results.

$v \: = \: \dfrac{me^{4}}{(4\pi \:)^{3} €^{2}_{0}( \dfrac{h}{2\pi})^{3}} ( \dfrac{1}{(n - 1) ^{2}} - \dfrac{1}{ {n}^{2}}$

$\dfrac{me^{4} (2n - 1)}{(4\pi \:)^{3}€^{2}_{0}( \dfrac{h}{2\pi})^{3} {n}^{2}(n - 1) ^{2}}$

For the large n we can write (2n-1) =~ 2n and (n-1) =~ n.

Therefore, v =

$\dfrac{me^{4}}{32\pi \: ^{3}€ ^{2}_{0}( \dfrac{h}{2\pi}) ^{3} n^{2}}$ Equation 4

Classical relation of frequency of revolution of an electron is given below as

$v_{e} = \dfrac{v}{2\pi \: r}$ Equation 5

Where,

• Velocity of the electron in $^{th}$ orbit us given as

$\dfrac{e^{2}}{4\pi \: €_{0}( \dfrac{h}{2\pi})n}$ Equation 6

v = Radius

$\dfrac{4\pi \: € _{0}( \dfrac{h}{2\pi}) ^{2}}{r \: = \: me ^{2} }n ^{2}$ Equation 7

Now putting the value of equation 6 and 7 in 8 we get the following results,

$v_{e} = \dfrac{me^{4}}{32\pi \: ^{3}€^{2}_{0}( \dfrac{h}{2\pi})^{3}n^{3}}$ Equation 8

Hence, the frequency of radiation emitted by the hydrogen atom is given ton it's classical orbit frequency.

Answer 2

Answer:

According to Bohr model

Explanation:

we can conclude this answer with the help of frequency formula v = Lamda /c

so at last the balmer series will bee helpful for this