Question 131: Atoms & Nuclei

Question

The smallest wavelength of Lyman series is 91 nm . The difference between the largest wavelengths of Paschen and Balmer series is nearly $\_\_\_\_$ nm.

Accepted Answer

The energy of $n^{	ext {th }}$ energy state of hydrogen atom is $E_n=-\frac{13.6}{n^2}$
The wavelengths of the spectral lines of hydrogen-like atoms are governed by the Rydberg formula
$$ \frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}ight) $$
Where :
$\lambda$ is the wavelength of the emitted photon.
R is the Rydberg constant.
$\mathrm{n}_1$ is the principal quantum number of the lower energy level.
$\mathrm{n}_2$ is the principal quantum number of the higher energy level ( $\mathrm{n}_2>\mathrm{n}_1$ ).
The smallest wavelength corresponds to the highest energy transition (from $\mathrm{n}_2=\infty$ to $\mathrm{n}_1$ ).
The largest wavelength corresponds to the lowest energy transition (from the immediate next energy level, $\mathrm{n}_2= \mathrm{n}_1+1$, down to $\mathrm{n}_1$ ).
For the Lyman series, the electron transitions fall to the ground state, so $\mathrm{n}_1=1$. The smallest wavelength $\left(\lambda_{\mathrm{L}(\min )}ight)$ occurs when $\mathrm{n}_2=\infty$.
It is given that $\lambda_{\mathrm{L}, \min }=91 \mathrm{~nm}$.
Using the Rydberg formula :
$$ \frac{1}{\lambda_{\mathrm{L}, \min }}=\mathrm{R}\left(\frac{1}{1^2}-\frac{1}{\infty^2}ight) $$
$$\Rightarrow $$ $$ \frac{1}{91}=\mathrm{R}(1-0) \Rightarrow \mathrm{R}=\frac{1}{91} \mathrm{~nm}^{-1} $$
For the Balmer series, the lower energy level is $\mathrm{n}_1=2$.
The largest wavelength ( $\lambda_{B, \max }$ corresponds to the smallest energy jump, which is from $\mathrm{n}_2=3$ to $\mathrm{n}_1=2$.
Applying the formula :
$$ \frac{1}{\lambda_{\mathrm{B}, \max }}=\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{3^2}ight)=\mathrm{R}\left(\frac{5}{36}ight) $$
$$\Rightarrow $$ $\frac{1}{\lambda_{\mathrm{B}, \max }}=\frac{1}{91} 	imes \frac{5}{36} \mathrm{~nm}^{-1}$
$$\Rightarrow $$ $$ \lambda_{\mathrm{B}, \max }=\frac{91 	imes 36}{5} \mathrm{~nm}=655.2 \mathrm{~nm} $$
For the Paschen series, the lower energy level is $\mathrm{n}_1=3$.
The largest wavelength ( $\lambda_{P, \max }$ ) corresponds to the smallest energy jump, which is from $\mathrm{n}_2=4$ to $\mathrm{n}_1=3$.
Applying the formula :
$$ \frac{1}{\lambda_{\mathrm{P}, \max }}=\mathrm{R}\left(\frac{1}{3^2}-\frac{1}{4^2}ight)=\mathrm{R}\left(\frac{1}{9}-\frac{1}{16}ight) $$
$$\Rightarrow $$ $\frac{1}{\lambda_{\mathrm{P}, \max }}=\frac{1}{91} 	imes \frac{7}{144} \mathrm{~nm}^{-1}$
$$ \Rightarrow \lambda_{P, \max }=\frac{91 	imes 144}{7} \mathrm{~nm}=1871 \mathrm{~nm} $$
So, the difference between these two largest wavelengths :
$$ \Delta \lambda=\lambda_{\mathrm{P}, \max }-\lambda_{\mathrm{B}, \max } $$
$$\Rightarrow $$ $\Delta \lambda=1872 \mathrm{~nm}-655.2 \mathrm{~nm}$
$$\Rightarrow $$ $$ \Delta \lambda=1216.8 \mathrm{~nm} \approx 1217 \mathrm{~nm} $$
Therefore, the difference between the largest wavelengths of Paschen and Balmer series is nearly 1217 nm.
Hence, the correct option is (C).

Answer

1784

Answer

1875

Answer

1217

Answer

1550

Atoms & Nuclei

The smallest wavelength of Lyman series is 91 nm . The difference between the largest wavelengths of Paschen and Balmer series is nearly \_\_\_\_ nm.

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