The shortest wavelength of the spectral lines in the Lyman series of hydrogen spectrum is 915\mathop A\limits^o . The longest wavelength of spectral lines in the Balmer series will be \underline\hspace2cm\mathop A\limits^o .

Atoms and Nuclei PYQ — JEE Main 2024 · 5 Apr · Shift 2 · Q87 with Solution

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Correct answer: 6588

Use the Rydberg formula

For hydrogen spectral lines,
$\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right), \qquad n_2 > n_1$
Shortest wavelength in Lyman series

In the Lyman series, transitions end at $n_1=1$ .

The shortest wavelength corresponds to the largest energy gap, i.e. $n_2 \to \infty$ .

So,
$\frac{1}{\lambda_{L,\min}} = R\left(1 - 0\right)=R$
Given,
$\lambda_{L,\min}=915\,\text{\AA}$
hence
$R = \frac{1}{915\,\text{\AA}}$
Longest wavelength in Balmer series

In the Balmer series, transitions end at $n_1=2$ .

The longest wavelength corresponds to the smallest energy gap, i.e. transition from $n_2=3$ to $n_1=2$ .

Therefore,
$\frac{1}{\lambda_{B,\max}} = R\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R\left(\frac{1}{4} - \frac{1}{9}\right) = R\cdot \frac{5}{36}$
Thus,
$\lambda_{B,\max} = \frac{36}{5R}$
Using $R=\frac{1}{915}$ ,
$\lambda_{B,\max} = \frac{36}{5}\times 915$
Calculate
$\lambda_{B,\max} = 7.2 \times 915 = 6588\,\text{\AA}$
Final answer
$\boxed{6588\,\text{\AA}}$
Comparison with stored correct answer

Stored correct answer = $6588$

This matches the derived answer.

View written solutionFree

Correct answer: 6588

Use the Rydberg formula

For hydrogen spectral lines,
$\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right), \qquad n_2 > n_1$
Shortest wavelength in Lyman series

In the Lyman series, transitions end at $n_1=1$ .

The shortest wavelength corresponds to the largest energy gap, i.e. $n_2 \to \infty$ .

So,
$\frac{1}{\lambda_{L,\min}} = R\left(1 - 0\right)=R$
Given,
$\lambda_{L,\min}=915\,\text{\AA}$
hence
$R = \frac{1}{915\,\text{\AA}}$
Longest wavelength in Balmer series

In the Balmer series, transitions end at $n_1=2$ .

The longest wavelength corresponds to the smallest energy gap, i.e. transition from $n_2=3$ to $n_1=2$ .

Therefore,
$\frac{1}{\lambda_{B,\max}} = R\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R\left(\frac{1}{4} - \frac{1}{9}\right) = R\cdot \frac{5}{36}$
Thus,
$\lambda_{B,\max} = \frac{36}{5R}$
Using $R=\frac{1}{915}$ ,
$\lambda_{B,\max} = \frac{36}{5}\times 915$
Calculate
$\lambda_{B,\max} = 7.2 \times 915 = 6588\,\text{\AA}$
Final answer
$\boxed{6588\,\text{\AA}}$
Comparison with stored correct answer

Stored correct answer = $6588$

This matches the derived answer.