Question

calculate the wavelength of the light emitted by the transition of electron from n2=3 in the balmer series​

Answer 1

Given

• An elctron moves from n = 3 to n = 2

To Find

• Wavelength

Solution

☯ 1/λ = RZ²(1/n²L - 1/n²H)

• Where R is the Rydberg's constant which is equal to 109667 cm⁻¹

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✭ According to the Question :

➞ 1/λ = RZ²(1/n²L - 1/n²H)

• nL = 2
• nH = 3
• And we may assume that it is on a Hydrogen atom whose atomic number if 1

➞ 1/λ = 109667 × 1²(1/2² - 1/3²)

➞ 1/λ = 109667 × (1/4 - 1/9)

➞ 1/λ = 109667 × [(9-4)/(9×4)]

➞ 1/λ = 109667 × (5/36)

➞ 1/λ = (109667 × 5)/36

➞ λ = 3/(109667 × 5)

➞ λ = 0.000065 cm⁻¹

∴ The wavelength is 0.000065 cm⁻¹

Answer 2

Given :

• The transition of electron from n2=3 in the balmer series.

What To Find :

• Wavelength of light

Solution :

The expression for the wavelength is

$\large\bf\blue{ \hookrightarrow} \rm \:\large \rm \: \frac{1}{λ} \: = \: R {Z}^{2} \: \bigg( \: \frac{1}{ {(n_1)}^{2}} \: - \: \frac{1}{ {(n_2)}^{2} } \bigg)$

Where ,

• λ= wavelength
• R is Rydberg's Constant.
• Z= Atomic number of atom
• Value of R is 109667 cm-¹
• n1 = 2
• n2 = 3

Let assume that it is on hydrogen atom and atomic number of hydrogen is 1

Substuting the values in above expression

$\begin{cases}\bf\red{ \implies} \rm \: \frac{1}{λ} \: = \: 109667 \times \: {(1)}^{2} \: \bigg( \: \frac{1}{ {(2)}^{2}} \: - \: \frac{1}{ {(3)}^{2} } \bigg) \\ \\ \bf\red{ \implies} \rm \: \frac{1}{λ} \: = \: 109667\: \bigg( \: \frac{1}{ {(4)}} \: - \: \frac{1}{ {(9)} } \bigg) \\ \\ \bf\red{ \implies} \rm \: \frac{1}{λ} \: = \: 109667\: \bigg( \frac{9 \: - \: 4}{36} \bigg) \\ \\ \bf\red{ \implies} \rm \: \frac{1}{λ} \: = \: 109667\: \bigg( \frac{5}{36} \bigg) \\ \\ \bf\red{ \implies} \rm \: \frac{1}{λ} \: = \frac{\: 109667\: \times \: 5 }{36} \\ \\ \bf\red{ \implies} \rm \: \frac{1}{λ} \: = \frac{\: 548335 }{36} \\ \\ \bf\red{ \implies} \rm \: λ \: = \frac{36}{548335} \\ \\ \bf\red{ \implies} \rm \: λ \: = \: \frac{ \cancel{36}}{ \cancel{548335}} \\ \\ \bf\red{ \implies} \rm \: λ \: = 0.00006565 \end{cases}$

$\boxed{\rm \: \rm\purple{ The \: wavelength \: is \: \bf\purple{ 0.00006565} \rm \: {cm}^{ - 1} }}$