Question

The shortest wavelength of the line in hydrogen atomic spectrum of Lyman series where R = 109678 cmrt is​

Answer 1

Solution:

For Hydrogen, Z = 1

For shortest wavelength in Lymann series, n₁ = 1, n₂ = ∞

Rydberg's Formula states that:

$\maltese \: \: \boxed{ \pink{ \bf{ \dfrac{1}{ \lambda} = R _{h} \times {z}^{2} \bigg( \dfrac{1}{(n_1)^{2} } - \dfrac{1}{(n_2)^{2} } \bigg) }}}$

where:

• λ is the wavelength
• Rh is the Rydberg constant = 109678
• Z is the atomic number
• n₁ = 1
• n₂ = ∞

Substitute the given values in the above equation.

$\sf{ \dfrac{1}{ \lambda} = 109678 \times {1}^{2} \times \bigg( \dfrac{1}{ {1}^{2} } - \dfrac{1}{ \infin ^{2} } \bigg) }$

$\dashrightarrow \: \: \sf{ \dfrac{1}{ \lambda} = (109678 \times 1) \times 1}$

$\dashrightarrow \: \: \sf{ \dfrac{1}{ \lambda} = 109678 \: {cm}^{ - 1} }$

$\dashrightarrow \: \: \sf{ \lambda = \dfrac{1}{109678 \: \: {cm}^{ - 1} } }$

$\dashrightarrow \: \: \sf{ \lambda = 9.117 \times 10^{ - 6} \: cm }$

$\dashrightarrow \: \: \sf{ \lambda = 911.7 \times {10}^{ - 10} \: m}$

$\dashrightarrow \: \: \sf{ \lambda = 911.7 Å}$

∴ Shortest wavelength of Atomic Hydrogen spectrum in Lyman series = 911.7 Å

Answer 2

Answer:

Solution:

For Hydrogen, Z = 1

\begin{gathered}\\\end{gathered}

For shortest wavelength in Lymann series, n₁ = 1, n₂ = ∞

\begin{gathered}\\\end{gathered}

Rydberg's Formula states that:

\begin{gathered} \maltese \: \: \boxed{ \pink{ \bf{ \dfrac{1}{ \lambda} = R _{h} \times {z}^{2} \bigg( \dfrac{1}{(n_1)^{2} } - \dfrac{1}{(n_2)^{2} } \bigg) }}} \\ \\ \end{gathered}

✠

λ

1

=R

h

×z

2

(

(n

1

)

2

1

−

(n

2

)

2

1

)

where:

λ is the wavelength

Rh is the Rydberg constant = 109678

Z is the atomic number

n₁ = 1

n₂ = ∞

\begin{gathered}\\\end{gathered}

Substitute the given values in the above equation.

\begin{gathered}\\\end{gathered}

\begin{gathered} \sf{ \dfrac{1}{ \lambda} = 109678 \times {1}^{2} \times \bigg( \dfrac{1}{ {1}^{2} } - \dfrac{1}{ \infin ^{2} } \bigg) } \\ \\ \end{gathered}

λ

1

=109678×1

2

×(

1

2

1

−

∞

2

1

)

\begin{gathered}\dashrightarrow \: \: \sf{ \dfrac{1}{ \lambda} = (109678 \times 1) \times 1} \\ \\ \end{gathered}

⇢

λ

1

=(109678×1)×1

\begin{gathered}\dashrightarrow \: \: \sf{ \dfrac{1}{ \lambda} = 109678 \: {cm}^{ - 1} } \\ \\ \end{gathered}

⇢

λ

1

=109678cm

−1

\begin{gathered} \dashrightarrow \: \: \sf{ \lambda = \dfrac{1}{109678 \: \: {cm}^{ - 1} } } \\ \\ \end{gathered}

⇢λ=

109678cm

−1

1

\begin{gathered}\dashrightarrow \: \: \sf{ \lambda = 9.117 \times 10^{ - 6} \: cm } \\ \\ \end{gathered}

⇢λ=9.117×10

−6

cm

\begin{gathered} \dashrightarrow \: \: \sf{ \lambda = 911.7 \times {10}^{ - 10} \: m} \\ \\ \end{gathered}

⇢λ=911.7×10

−10

m

\begin{gathered} \dashrightarrow \: \: \sf{ \lambda = 911.7 Å} \\ \\ \end{gathered}

⇢λ=911.7

A

˚

∴ Shortest wavelength of Atomic Hydrogen spectrum in Lyman series = 911.7 Å