a) How are the characteristic X-rays spectrum formed? b) If the value of Rydberg constant of hydrogen is 10973

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Q. a) How are the characteristic X-rays spectrum formed? b) If the value of Rydberg constant of hydrogen is 109737 cm^-1, determine the longest and shortest wavelengths of the Balmer series. OR

a) What is mean life of a radioactive element? b) Establish a relation between half-life and decay constant.

West Bengal WbchseWest Bengal HS (WBCHSE) Board 2019Subjective· 3mImportance★★★★★

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Concept understanding — Atomic Spectra & the Hydrogen Spectrum

Atomic Spectra & the Hydrogen Spectrum

The big idea. When an excited electron drops from a higher level n₂ to a lower level n₁, the atom emits a photon whose energy is exactly the gap between the levels. Because the levels are quantized, only certain photon energies — and therefore only certain wavelengths — appear: a line spectrum, the atom's fingerprint. One equation, the Rydberg formula, delivers the wavelength of every line.

The Rydberg formula

For a hydrogen-like species of nuclear charge Z:

ν̄ = 1/λ = R_H · Z² · (1/n₁² − 1/n₂²), with n₁ < n₂ and R_H = 1.097×10⁷ m⁻¹ (= 109677 cm⁻¹).

ν̄ is the wavenumber (m⁻¹ or cm⁻¹), λ the wavelength, and the frequency is ν = c/λ = c·ν̄.
n₁ is the lower level (where the electron lands), n₂ the upper level (where it starts). Keep 1/n₁² − 1/n₂² positive.
Everything scales as Z²: a He⁺ line (Z = 2) sits at ¼ the wavelength of the same hydrogen transition.

The spectral series

Each series is defined by the level the electron falls to (n₁):

Series	n₁	Region
Lyman	1	Ultraviolet
Balmer	2	Visible
Paschen	3	Infrared
Brackett	4	Infrared
Pfund	5	Infrared

Only the Balmer series lies in the visible region — that is why it was discovered first.

Longest and shortest wavelength in a series

Within one series (fixed n₁):

The first line (n₂ = n₁+1) has the smallest energy gap, hence the longest wavelength.
The series limit (n₂ → ∞) has the largest gap, hence the shortest wavelength: 1/λ_min = R_H·Z²/n₁².

A frequent trap is swapping these — remember longest λ ↔ smallest 1/λ ↔ smallest energy jump.

Counting the lines

If a single electron is in level n and cascades down to the ground state, or a large sample of atoms is excited to level n, the number of distinct spectral lines is

N = n(n−1)/2.

Between two arbitrary levels n₂ and n₁ the count is (n₂−n₁)(n₂−n₁+1)/2. (The common wrong formula n(n+1)/2 counts a non-existent n→n "line".)

Energies add, wavelengths do not

For consecutive transitions the energies (and wavenumbers) add, never the wavelengths:

ΔE(n₃→n₁) = ΔE(n₃→n₂) + ΔE(n₂→n₁), so 1/λ₃₁ = 1/λ₃₂ + 1/λ₂₁. …

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a) Characteristic X-rays arise from transitions of electrons between inner atomic shells after a high-energy electron knocks out an inner-shell electron. b) For the Balmer series, the longest wavelength is ≈ 6561 Å (3→2 transition) and the shortest is ≈ 3646 Å (∞→2 transition).

a) Formation of characteristic X-ray spectrum: When high-energy electrons strike a metal target, they can knock out an electron from an inner shell (say the K-shell) of a target atom, creating a vacancy. An electron from a higher shell (L, M, ...) then falls into this vacancy, and the energy difference between the two shells is emitted as an X-ray photon of a definite (characteristic) energy/wavelength, specific to the target element. These transitions give rise to the $K_\alpha, K_\beta$ , etc., lines superimposed on the continuous (bremsstrahlung) X-ray spectrum.

b) Balmer series wavelengths: The Balmer series corresponds to transitions to $n_1=2$ from $n_2=3,4,5,\dots$

$\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right), \quad R = 109737\ \text{cm}^{-1}$

Longest wavelength (smallest energy gap): $n_2=3$ : …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set ANNUAL1 markMCQ
$Q. The electron of a hydrogen atom is excited to nth level. How many total possible spectral lines will be found for the transition from this excited state to all other states below n?$
(a) n
(b) n(n−1)/2
(c) n(n+1)/2
(d) (n+1)/2
›Reveal solutionSolution

The number of possible transition lines from level n down to all lower levels is the number of ways to pick 2 levels from n, i.e. n(n−1)/2.

From the nth excited level, an electron can jump to any of the (n−1) lower levels directly, or via intermediate levels — every distinct pair of levels (upper, lower) among the n levels (1 throug …

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CBSE 2023Set ANNUAL1 markMCQ
$Q. In hydrogen spectrum the ratio of the minimum wavelengths of Lyman and Balmer series is$
(a) 1 : 4
(b) 4 : 1
(c) 5 : 4
(d) 4 : 3
›Reveal solutionSolution

The minimum (series-limit) wavelength of each series comes from n → ∞; computing both from the Rydberg formula gives a ratio of 1:4.

The Rydberg formula for hydrogen is $\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$ . The minimum wavelength of a series occurs for the largest possible transition energy, i.e. $n_2 \to \infty$ .

Lyman series ( $n_1 = 1$ ):

$\dfrac{1}{\lambda_{Ly,min}} = R\left(1 - 0\right) = R \implies \lambda_{Ly,min} = \dfrac{1}{R}$

Balmer series ( $n_1 = 2$ ): …

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CBSE 2022Set ANNUAL1 markMCQ
$Q. Which of the following transitions in hydrogen atom emits a photon of lowest frequency?$
(a) n = 2 to n = 1
(b) n = 4 to n = 3
(c) n = 3 to n = 1
(d) n = 4 to n = 2 OR The density of nucleus in kg/m³ is of the order of
(a) 10¹¹
(b) 10¹⁴
(c) 10¹⁷
(d) 10²⁰
›Reveal solutionSolution
Photon frequency $\propto$ energy difference $\propto \left(\frac1{n_1^2}-\frac1{n_2^2}\right)$ ; comparing all four options, $n=4\to3$ gives the smallest value.

For a hydrogen-atom transition from level $n_2$ to $n_1$ ( $n_2>n_1$ ), the emitted photon's energy (hence frequency, since $E=h\nu$ ) is

$E \propto \frac{1}{n_1^2}-\frac{1}{n_2^2}$

Computing this factor for each option:

$n=2\to1$ : $1-\frac14 = 0.75$ …
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CBSE 2016Set ANNUAL1 markMCQ
$Q. The ratio of minimum wavelength of Lyman and Balmer series in hydrogen spectrum will be$
(a) 10
(b) 5
(c) 0.25
(d) 1.25
›Reveal solutionSolution

The minimum (series-limit) wavelengths of the Lyman and Balmer series are in the ratio 1:4, i.e. 0.25.

Using the Rydberg formula 1/λ = R(1/n1² − 1/n2²): the minimum wavelength of a series corresponds to the electron transition from n2 = ∞.

Lyman series (n1 = 1): 1/λ_L(min) = R(1 − 0) = R ⟹ λ_L(min) = 1/R

…

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Q.a) How are the characteristic X-rays spectrum formed? b) If the value of Rydberg constant of hydrogen is 109737 cm^-1, determine the longest and shortest wavelengths of the Balmer series. OR

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