The shortest wavelength of the spectral lines in the Lyman series of hydrogen spectrum is 915 Å. The longest wavelength of spectral lines in the Balmer series will be _______ Å.
Correct Answer: 6588
Approach Solution - 1
To find the longest wavelength of the spectral lines in the Balmer series, we begin by using the relationship for the energy levels of hydrogen and the transitions involved.
Step 1: Lyman Series
For the Lyman series:
\[ \frac{hc}{\lambda} = -13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \quad (\text{in eV}), \]
where \( n_1 = 1 \) for the Lyman series and \( n_2 \) varies.
Given the shortest wavelength in the Lyman series:
\[ \lambda_{\text{Lyman}} = 915 \, \text{\AA}. \]
Step 2: Balmer Series
For the Balmer series:
- \( n_1 = 2 \)
- \( n_2 = 3 \) for the longest wavelength transition.
The energy difference for the transition is given by:
\[ \frac{hc}{\lambda_1} = -13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right). \]
Calculating:
\[ \frac{hc}{\lambda_1} = -13.6 \left( \frac{1}{4} - \frac{1}{9} \right). \]
Simplifying:
\[ \frac{hc}{\lambda_1} = -13.6 \left( \frac{5}{36} \right). \]
Step 3: Relating the Wavelengths
Using the given data for the Lyman series:
\[ \lambda_1 = \lambda_{\text{Lyman}} \times \frac{36}{5}. \]
Substituting the given value:
\[ \lambda_1 = 915 \times \frac{36}{5} = 6588 \, \text{\AA}. \]
Therefore, the longest wavelength of spectral lines in the Balmer series is \( 6588 \, \text{\AA} \).
Approach Solution -2
Shortest wavelength in Lyman series (λL) = 915 Å
We need to find the longest wavelength in Balmer series (λB).
Step 2: Formula for wavelength of hydrogen spectral lines.
According to Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where R = 1.097 × 10⁷ m⁻¹ (Rydberg constant).
Step 3: For Lyman series (n₁ = 1).
Shortest wavelength → n₂ = ∞
\[ \frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R \] \[ \lambda_L = \frac{1}{R} \] Given λL = 915 Å = 9.15 × 10⁻⁸ m.
Hence, \[ R = \frac{1}{9.15 \times 10^{-8}} = 1.093 \times 10^7 \, \text{m}^{-1} \] which matches the standard Rydberg value.
Step 4: For Balmer series (n₁ = 2).
Longest wavelength → n₂ = 3.
\[ \frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \] \[ \frac{1}{\lambda_B} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right) \] \[ \lambda_B = \frac{36}{5R} \] Substitute \( R = 1.093 \times 10^7 \, \text{m}^{-1} \):
\[ \lambda_B = \frac{36}{5 \times 1.093 \times 10^7} = 6.588 \times 10^{-7} \, \text{m} = 6588 \, \text{Å} \]
Step 5: Final Answer.
The longest wavelength of spectral lines in the Balmer series is:
\[ \boxed{6588 \, \text{Å}} \]
Final Answer: 6588 Å
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Atomic Physics Questions
- When a hydrogen atom going from \( n = 2 \) to \( n = 1 \) emits a photon, its recoil speed is \( \frac{x}{5} \) m/s. Where \( x = \, \) _____. (Use: mass of hydrogen atom \( = 1.6 \times 10^{-27} \, \text{kg} \))
- The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly :
- A hydrogen atom changes its state from \( n = 3 \) to \( n = 2 \). Due to recoil, the percentage change in the wavelength of emitted light is approximately \( 1 \times 10^{-n} \). The value of \( n \) is ______. \([ \text{Given: } Rhc = 13.6 \, \text{eV}, \, hc = 1242 \, \text{eV nm}, \, h = 6.6 \times 10^{-34} \, \text{J s}, \, \text{mass of the hydrogen atom} = 1.6 \times 10^{-27} \, \text{kg} ]\)
- The angular momentum of an electron in a hydrogen atom is proportional to: (Where \( r \) is the radius of the orbit of the electron)
- A particular hydrogen-like ion emits the radiation of frequency $3 \times 10^{15}$ Hz when it makes a transition from $n = 2$ to $n = 1$. The frequency of radiation emitted in the transition from $n = 3$ to $n = 1$ is \[\frac{x}{9} \times 10^{15} \, \text{Hz}, \, \text{when} \, x = _____ .]
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.