Hydrogen atom is bombarded with electrons accelerated through a potential difference of V$V$, which causes excitation of hydrogen atoms. If the experiment is being performed at T = 0 K$T = 0\text{ K}$, the minimum potential difference needed to observe any Balmer series lines in the emission spectra will be (α)/(10) V$\frac{\alpha}{10}\text{ V}$, where α =$\alpha =$ ________.
Numerical Answer Type:
Enter a numerical valueAnswer: 121 to 121+4 marks
Solution & Explanation
Related Formula
The energy of an electron in the n$n$-th state of a hydrogen atom is:
Balmer series emission requires excitation of the electron to at least n=3$n=3$. Ground state n=1$n=1$ to n=3$n=3$ excitation energy is 12.09 eV ≈ 12.1 eV$12.09\text{ eV} \approx 12.1\text{ eV}$, matching α = 121$\alpha = 121$ when multiplied by 10$10$.
Bohr orbital dimensions scale quadratically with quantum index (n²$n^2$), rendering quick ratios straightforward via simple squaring operations.
Chapter Mix
Class 12 Physics: Atoms
Qjee_main_2024_29_jan_morningBohr Model of Hydrogen Atom
When a hydrogen atom going from n = 2$n = 2$ to n = 1$n = 1$ emits a photon, its recoil speed is (x)/(5) ~m / s$\frac{x}{5} \mathrm{~m / s}$. Where x =$x =$ _______. (Use: mass of hydrogen atom = 1.6 × 10⁻²⁷ ~kg$= 1.6 \times 10^{-27} \mathrm{~kg}$)
Numerical Answer.Answer: 17 to 17
Solution
Related Formula
By conservation of linear momentum, the momentum of the recoiling hydrogen atom matches the momentum of the emitted photon:
patom = pphoton m v = (Δ E)/(c)$$p_{\text{atom}} = p_{\text{photon}} \implies m v = \frac{\Delta E}{c}$$
Hence, the recoil speed v$v$ is:
v = (Δ E)/(m c)$$v = \frac{\Delta E}{m c}$$
where Δ E$\Delta E$ is the energy difference between the electronic energy levels.
Core Logic
For a hydrogen transition from n = 2$n = 2$ to n = 1$n = 1$:
Equating the computed speed to the given parameter form:
3.4 = (x)/(5) x = 3.4 × 5 = 17$$3.4 = \frac{x}{5} \implies x = 3.4 \times 5 = 17$$
Therefore, the value of x$x$ is 17$17$.
Pattern Recognition
Always look for clean component cancellation matches prior to raw exponential expansion. In this problem, the value 1.6$1.6$ in the electron charge value cancels with the structural atomic mass constant value perfectly.
Chapter Mix
Class 12 Physics: Atoms
Q37jee_main_2024_30_january_eveningBohr Model Magnetic Moment
An electron revolving in nth$n^{\text{th}}$ Bohr orbit has magnetic moment μₙ$\mu_{\mathrm{n}}$. If μₙ ∝ nx$\mu_{\mathrm{n}} \propto n^{\mathrm{x}}$, the value of x$\mathbf{x}$ is:
A.2$2$
B.1$1$
C.3$3$
D.0$0$
Solution
Related Formula
μₙ = i A = (e)/(T) π r²$$\mu_n = i A = \frac{e}{T} \pi r^2$$
Alternative relation using angular momentum:
μₙ = (e)/(2m) L$$\mu_n = \frac{e}{2m} L$$
Core Logic
From Bohr's quantization condition, angular momentum L$L$ is given by:
L = (nh)/(2π)$$L = \frac{nh}{2\pi}$$
Therefore, the magnetic moment can be written directly as:
We can also derive this via velocity and radius:
v ∝ (1)/(n)$v \propto \frac{1}{n}$ and r ∝ n²$r \propto n^2$.
μₙ = (e v r)/(2) ∝ ((1)/(n)) (n²) = n$$\mu_n = \frac{e v r}{2} \propto \left(\frac{1}{n}\right) (n^2) = n$$
Comparing with μₙ ∝ n^x$\mu_n \propto n^x$, we find x = 1$x = 1$.
Pattern Recognition
Magnetic moment of an electron in a Bohr orbit is always directly proportional to the principal quantum number n$n$. It is an integer multiple of the Bohr magneton (μB = (eh)/(4π m)$\mu_B = \frac{eh}{4\pi m}$).
Chapter Mix
Class 12 Physics: Atoms
Q36jee_main_2024_30_jan_morningBohr Model and Electron Energy
The ratio of the magnitude of the kinetic energy to the potential energy of an electron in the 5th$5^{\text{th}}$ excited state of a hydrogen atom is :
A.4$4$
B.(1)/(4)$\frac{1}{4}$
C.(1)/(2)$\frac{1}{2}$
D.1$1$
Solution
Related Formula
KE = -E$\text{KE} = -E$PE = 2E$\text{PE} = 2E$
KE = (1)/(2) |PE|$$\text{KE} = \frac{1}{2} |\text{PE}|$$
Core Logic
In any allowed Bohr orbit (for any value of principal quantum number n$n$), the relationship between kinetic energy (KE), potential energy (PE), and total energy (E) strictly obeys the virial theorem for a Coulombic force field.
|PE| = 2 × KE$$|\text{PE}| = 2 \times \text{KE}$$
Step 1: Formulate the Ratio
The question asks for the ratio of the magnitude of KE to the magnitude of PE.
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