Question 130: Atoms & Nuclei

Question

In hydrogen atom spectrum, ( $R \rightarrow$ Rydberg's constant)
A. the maximum wavelength of the radiation of Lyman series is $\frac{4}{3 R}$
B. the Balmer series lies in the visible region of the spectrum
C. the minimum wavelength of the radiation of Paschen series is $\frac{9}{R}$
D. the minimum wavelength of Lyman series is $\frac{5}{4 R}$
Choose the correct answer from the options given below :

Accepted Answer

Using the Rydberg formula :
$$ \frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right) $$
Where, $\lambda$ is the wavelength of radiated photon when electron makes a transition from initial energy state $n_2$ to final energy state $\mathrm{n}_1$, and R is the Rydberg's constant.
For statement A
The given series is Lyman ( $\mathrm{n}_1=1$ ).
Maximum wavelength implies minimum energy. The transition corresponds to the smallest energy gap, which is between the first excited state ( $\mathrm{n}_2=2$ ) and the ground state ( $\mathrm{n}_1=1$ ).
$$ \frac{1}{\lambda_{\max }}=\mathrm{R}\left(\frac{1}{1^2}-\frac{1}{2^2}\right) $$
$$\Rightarrow $$ $$ \frac{1}{\lambda_{\max }}=\mathrm{R}\left(1-\frac{1}{4}\right)=\frac{3 \mathrm{R}}{4} \Rightarrow \lambda_{\max }=\frac{4}{3 \mathrm{R}} $$
Hence, the statement A is true.
For statement B
The given series is Balmer ( $\mathrm{n}_1=2$ ).
The Balmer series corresponds to electron transitions from higher shells ( $n=3,4,5 \ldots$ ) to the second shell $(\mathrm{n}=2)$. The wavelengths produced (e.g., $656 \mathrm{~nm}, 486 \mathrm{~nm})$ fall within the range of approximately 400 nm to 700 nm , which is the visible spectrum
So, Balmer series lines are indeed visible. Hence, the statement B is true.
For Statement C
The given series is Paschen ( $\mathrm{n}_1=3$ ).
Minimum wavelength implies maximum energy. The transition corresponds to the largest energy gap, which is from infinity $\left(\mathrm{n}_2=\infty\right)$ to the level $\mathrm{n}_1=3$.
$$ \frac{1}{\lambda_{\min }}=R\left(\frac{1}{3^2}-\frac{1}{\infty^2}\right) $$
$$\Rightarrow $$ $$ \frac{1}{\lambda_{\min }}=\mathrm{R}\left(\frac{1}{9}\right)=\frac{\mathrm{R}}{9} \Rightarrow \lambda_{\min }=\frac{9}{\mathrm{R}} $$
Hence, the statement C is also true.
For statement D
The given series is Lyman ( $\mathrm{n}_1=1$ ).
Minimum wavelength means ( $\mathrm{n}_2=\infty \rightarrow \mathrm{n}_1=1$ ).
$$ \frac{1}{\lambda_{\min }}=\mathrm{R}\left(\frac{1}{1^2}-\frac{1}{\infty^2}\right) $$
$$\Rightarrow $$ $\frac{1}{\lambda_{\min }}=\mathrm{R}(1-0)=\mathrm{R}$
$$\Rightarrow $$ $$ \lambda_{\min }=1 \mathrm{R} $$
Hence, the statement D is false.
Statement A is Correct.
Statement B is Correct.
Statement C is Correct.
Statement D is Incorrect.
Therefore, the correct option is (D).

Answer

A, B Only

Answer

B, D Only

Answer

A, B and D Only

Answer

A, B and C Only

Atoms & Nuclei

Choose the correct answer: