Question:
In hydrogen spectrum, the frequency of the spectral line corresponding to electron transition $n_2 = 3$ to $n_1 = 2$ is $x$ Hz. What is the frequency (in Hz) of the spectral line corresponding to electron transition $n_2 = 4$ to $n_1 = 3$ of $\text{He}^+$ spectrum?
In hydrogen spectrum, the frequency of the spectral line corresponding to electron transition $n_2 = 3$ to $n_1 = 2$ is $x$ Hz. What is the frequency (in Hz) of the spectral line corresponding to electron transition $n_2 = 4$ to $n_1 = 3$ of $\text{He}^+$ spectrum?
Show Hint
Always write out the ratio of frequencies $\frac{\nu_1}{\nu_2} = \frac{Z_1^2}{Z_2^2} \times \frac{(1/n_{1a}^2 - 1/n_{2a}^2)}{(1/n_{1b}^2 - 1/n_{2b}^2)}$ to directly eliminate constants.
This reduces algebraic calculations and avoids units errors.
This reduces algebraic calculations and avoids units errors.
Updated On: Jul 22, 2026
- $\frac{5x}{7}$
- $\frac{7x}{5}$
- $\frac{20x}{7}$
- $\frac{7x}{20}$
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
This chemistry problem belongs to atomic structure.
We need to find the relationship between the frequency of a transition in a hydrogen atom and a transition in a helium ion ($\text{He}^+$).
Step 2: Key Formula or Approach:
The frequency $\nu$ of the emitted radiation during an electronic transition in a hydrogen-like species is given by the Bohr-Rydberg equation:
\[ \nu = c R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where $R$ is the Rydberg constant, $Z$ is the atomic number, $c$ is the speed of light, and $n_1, n_2$ are the principal quantum numbers of the lower and higher energy levels.
Step 3: Detailed Explanation:
• For the hydrogen atom ($Z = 1$):
The transition is from $n_2 = 3$ to $n_1 = 2$.
The frequency is $x$ Hz:
\[ x = c R (1)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \] \[ x = c R \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5}{36} c R \quad \text{--- (Equation 1)} \]
• For the Helium ion $\text{He}^+$ ($Z = 2$):
The transition is from $n_2 = 4$ to $n_1 = 3$.
Let the frequency of this transition be $\nu$:
\[ \nu = c R (2)^2 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \] \[ \nu = 4 c R \left( \frac{1}{9} - \frac{1}{16} \right) \] \[ \nu = 4 c R \left( \frac{16 - 9}{144} \right) \] \[ \nu = 4 c R \left( \frac{7}{144} \right) = \frac{7}{36} c R \quad \text{--- (Equation 2)} \]
• Dividing Equation 2 by Equation 1 to find the relation between $\nu$ and $x$:
\[ \frac{\nu}{x} = \frac{\frac{7}{36} c R}{\frac{5}{36} c R} = \frac{7}{5} \] \[ \nu = \frac{7x}{5} \]
Step 4: Final Answer:
The frequency of the transition for $\text{He}^+$ is $\frac{7x}{5}$.
This chemistry problem belongs to atomic structure.
We need to find the relationship between the frequency of a transition in a hydrogen atom and a transition in a helium ion ($\text{He}^+$).
Step 2: Key Formula or Approach:
The frequency $\nu$ of the emitted radiation during an electronic transition in a hydrogen-like species is given by the Bohr-Rydberg equation:
\[ \nu = c R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where $R$ is the Rydberg constant, $Z$ is the atomic number, $c$ is the speed of light, and $n_1, n_2$ are the principal quantum numbers of the lower and higher energy levels.
Step 3: Detailed Explanation:
• For the hydrogen atom ($Z = 1$):
The transition is from $n_2 = 3$ to $n_1 = 2$.
The frequency is $x$ Hz:
\[ x = c R (1)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \] \[ x = c R \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5}{36} c R \quad \text{--- (Equation 1)} \]
• For the Helium ion $\text{He}^+$ ($Z = 2$):
The transition is from $n_2 = 4$ to $n_1 = 3$.
Let the frequency of this transition be $\nu$:
\[ \nu = c R (2)^2 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \] \[ \nu = 4 c R \left( \frac{1}{9} - \frac{1}{16} \right) \] \[ \nu = 4 c R \left( \frac{16 - 9}{144} \right) \] \[ \nu = 4 c R \left( \frac{7}{144} \right) = \frac{7}{36} c R \quad \text{--- (Equation 2)} \]
• Dividing Equation 2 by Equation 1 to find the relation between $\nu$ and $x$:
\[ \frac{\nu}{x} = \frac{\frac{7}{36} c R}{\frac{5}{36} c R} = \frac{7}{5} \] \[ \nu = \frac{7x}{5} \]
Step 4: Final Answer:
The frequency of the transition for $\text{He}^+$ is $\frac{7x}{5}$.
Was this answer helpful?
0
0
Top AP EAPCET Chemistry Questions
- The four quantum numbers of the valence electron of potassium are :
- 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
- In which of the following the election gain enthalpy of elements is correctly arranged?
- The radii of $ \text{F,}{{\text{F}}^{-}}\text{O} $ and $ {{\text{O}}^{2-}} $ are in the order
- Which of the following has the highest first ionization potential?
View More Questions
Top AP EAPCET Bohr Model Of The Hydrogen Atom Questions
- The radius of H atom in its ground state is \(x~\text{pm}\). What is the ratio of radii in ground and \(n = 2\) states of He\(^+\)?
- The total number of electrons with magnetic quantum number \(m_l = 0\) in Fe\(^{3+}\) and Cr\(^+\) is respectively
- A spectral line of Lyman series of H-atom has a frequency of \(2.466 \times 10^{15} \, {s}^{-1}\). What is the transition responsible for this spectral line? (Rydberg constant \(R = 1.096 \times 10^7 \, {m}^{-1}\))
- The radius in the hydrogen atom in the ground state is $0.53$ \AA. The radius of $Li^{+2}$ ion (atomic number $= 3$) in a similar state is \ldots..
- The energy of the second bohr orbit of the hydrogen atom is $-328 \, \text{kJ mol}^{-1}$; hence the energy of the fourth bohr orbit would be:
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