The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes n = 2 → n = 3 and n = 4 → n = 6 transitions, respectively, is

The correct option is 1/4.

To find the ratio of the wavelengths of the light absorbed by a Hydrogen atom during the given transitions, we can use the Rydberg formula for the wavelength of light absorbed or emitted:
$\frac{1}{\lambda }=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
where:
• $\lambda$ is the wavelength of the light,
• $R_H$ is the Rydberg constant for Hydrogen,
• $n_1$ and $n_2$ are the principal quantum numbers of the energy levels (with $n_1<n_2$).

Step 1: Calculate the wavelength (${\lambda }_1$) for the transition $n=2\to n=3$
Here, $n_1=2$ and $n_2=3$.
Substituting these values into the Rydberg formula:
$\frac{1}{{\lambda }_1}=R_H\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$
$\frac{1}{{\lambda }_1}=R_H\left(\frac{1}{4}-\frac{1}{9}\right)$
$\frac{1}{{\lambda }_1}=R_H\left(\frac{9-4}{36}\right)$
$\frac{1}{{\lambda }_1}=R_H\left(\frac{5}{36}\right)$
Thus, the wavelength ${\lambda }_1$ is:
${\lambda }_1=\frac{36}{5R_H}$

Step 2: Calculate the wavelength (${\lambda }_2$) for the transition $n=4\to n=6$
Here, $n_1=4$ and $n_2=6$.
Substituting these values into the Rydberg formula:
$\frac{1}{{\lambda }_2}=R_H\left(\frac{1}{4^2}-\frac{1}{6^2}\right)$
$\frac{1}{{\lambda }_2}=R_H\left(\frac{1}{16}-\frac{1}{36}\right)$
Taking the least common multiple of 16 and 36, which is 144:
$\frac{1}{{\lambda }_2}=R_H\left(\frac{9-4}{144}\right)$
$\frac{1}{{\lambda }_2}=R_H\left(\frac{5}{144}\right)$
Thus, the wavelength ${\lambda }_2$ is:
${\lambda }_2=\frac{144}{\mathrm{5<!--\; <msub-->\; <mi>R</mi>\; <mi>H</mi>}}$

Step 3: Find the ratio of the wavelengths (${\lambda }_1/{\lambda }_2$)
Dividing ${\lambda }_1$ by ${\lambda }_2$:
$\frac{{\lambda }_1}{{\lambda }_2}=\frac{\left(\frac{36}{5R_H}\right)}{\left(\frac{144}{\mathrm{5<!--\; <msub-->\; <mi>R</mi>\; <mi>H</mi>}}\right)}$
Canceling the common factor of $\frac{5}{R_H}$ in the denominators:
$\frac{{\lambda }_1}{{\lambda }_2}=\frac{36}{144}$
Simplifying the fraction:
$\frac{{\lambda }_1}{{\lambda }_2}=\frac{1}{4}$

Therefore, the ratio of the wavelengths of the absorbed light is 1/4.