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Question: The ratio of wavelength of the last line of Balmer series and the last line of Lyman series is: A)...

The ratio of wavelength of the last line of Balmer series and the last line of Lyman series is:
A) 0.5
B) 2
C) 1
D) 4

Explanation

Solution

The emission spectrum of the hydrogen atom is classified in five series: Lyman, Balmer, Paschan, Bracket and fund. The Lyman series comes under the ultraviolet region and the remaining series are under infrared.

Complete step by step solution:
According to the question it is given that there are two series wavelengths, the last line of Balmer Series and Lyman Series.
The expression for emission spectrum is written as shown below.
⇒1λ=RZ2(1p2−1n2)\Rightarrow \dfrac{1}{\lambda } = R{Z^2}\left( {\dfrac{1}{{{p^2}}} - \dfrac{1}{{{n^2}}}} \right)

Where, λ\lambda is the wavelength, RR is Rydberg constant.
For Hydrogen atom, We know that Z=1Z = 1,
⇒1λ=R(1p2−1n2)\Rightarrow \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{p^2}}} - \dfrac{1}{{{n^2}}}} \right)

For Lyman series, We know that p=1p = 1, n=∞n = \infty ,
So, the emission spectrum is written as shown below.
⇒1λL=R(112−1∞2) ⇒1λL=R  \Rightarrow \dfrac{1}{{{\lambda _L}}} = R\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{\infty ^2}}}} \right) \\\ \Rightarrow \dfrac{1}{{{\lambda _L}}} = R \\\
This is the equation for the Lyman series as given above.

For Balmer series, We know that p=2p = 2, n=∞n = \infty
So, the emission spectrum is written as shown below.
⇒1λB=R(122−1∞2) ⇒1λB=R4  \Rightarrow \dfrac{1}{{{\lambda _B}}} = R\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{\infty ^2}}}} \right) \\\ \Rightarrow \dfrac{1}{{{\lambda _B}}} = \dfrac{R}{4} \\\
This is the equation for the Balmer series.

Take the ratio of both the equation of Lyman and Balmer series and it is written as shown below.

& \Rightarrow \dfrac{\dfrac{1}{{{\lambda }_{L}}}}{\dfrac{1}{{{\lambda }_{B}}}}=\dfrac{R}{R/4} \\\ & \therefore \dfrac{{{\lambda }_{B}}}{{{\lambda }_{L}}}=4 \\\ \end{aligned}$$ So, from the above calculation it is concluded that the ratio of wavelength of the last line of the Balmer series and the last line of the Lyman series is $4$. **Hence, the option (D) is correct.** **Note:** The wavelength of the Balmer and Lyman series can be calculated using the emission spectrum expression. The value of the variable p and n are different for all series. For Lyman series, we know that $p = 1$, $n = \infty $ For Balmer series, we know that $p = 2$, $n = \infty $ For Paschen series, we know that $p = 3$, $n = \infty $ For Bracket series, we know that $p = 4$, $n = \infty $ For Fund series, we know that $p = 5$, $n = \infty $