Question:
Let \(\lambda_L\) and \(\lambda_S\) be the longest and shortest wavelength photons in the Balmer series respectively. Find \(\frac{\lambda_L}{\lambda_S}\):
Let \(\lambda_L\) and \(\lambda_S\) be the longest and shortest wavelength photons in the Balmer series respectively. Find \(\frac{\lambda_L}{\lambda_S}\):
Show Hint
In hydrogen spectrum, longest wavelength corresponds to smallest energy transition.
Updated On: Jun 19, 2026
- \( \frac{9}{5} \)
- 2
- \( \frac{5}{2} \)
- \( \frac{8}{3} \)
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Use Balmer series formula.
\[ \frac{1}{\lambda} = R \left(\frac{1}{2^2} - \frac{1}{n^2}\right) \]
Step 2: Identify wavelengths.
- Longest wavelength: \(n=3 \to 2\) transition - Shortest wavelength: \(n=\infty \to 2\) transition
Step 3: Compute longest wavelength.
\[ \frac{1}{\lambda_L} = R\left(\frac{1}{4} - \frac{1}{9}\right) = R\left(\frac{5}{36}\right) \]
Step 4: Compute shortest wavelength.
\[ \frac{1}{\lambda_S} = R\left(\frac{1}{4} - 0\right) = \frac{R}{4} \]
Step 5: Take ratio.
\[ \frac{\lambda_L}{\lambda_S} = \frac{4}{5/36} = \frac{144}{5} \times \frac{1}{16} = \frac{9}{5} \]
Step 6: Final conclusion.
\[ \boxed{\frac{9}{5}} \]
\[ \frac{1}{\lambda} = R \left(\frac{1}{2^2} - \frac{1}{n^2}\right) \]
Step 2: Identify wavelengths.
- Longest wavelength: \(n=3 \to 2\) transition - Shortest wavelength: \(n=\infty \to 2\) transition
Step 3: Compute longest wavelength.
\[ \frac{1}{\lambda_L} = R\left(\frac{1}{4} - \frac{1}{9}\right) = R\left(\frac{5}{36}\right) \]
Step 4: Compute shortest wavelength.
\[ \frac{1}{\lambda_S} = R\left(\frac{1}{4} - 0\right) = \frac{R}{4} \]
Step 5: Take ratio.
\[ \frac{\lambda_L}{\lambda_S} = \frac{4}{5/36} = \frac{144}{5} \times \frac{1}{16} = \frac{9}{5} \]
Step 6: Final conclusion.
\[ \boxed{\frac{9}{5}} \]
Was this answer helpful?
0
0
Top AP EAPCET Physics Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is $ 0{}^\circ $ ) is filled in it, the level difference of the liquid in the two arms is h. The surface tension of the liquid is: (g = acceleration due to gravity)
- A rigid metallic sphere is spinning around its own axis in the absence of external torque. If the temperature is raised, its volume increases by $9 \%$. The change in its angular speed is
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions