Q.A hypothetical electromagnetic wave is shown in the figure below. Find out the wavelength of the radiation.

Showing the 12 most recent of 132 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.Match the following

   [!FORMULA] $\begin{array}{|c|c|} \hline \text{List-1 (Element)} & \text{List-2 (Block)} \\ \hline \text{A Hf} & \text{I s-block} \\ \text{B Ra} & \text{II p-Block} \\ \text{C Am} & \text{III d-Block} \\ \text{D At} & \text{IV f-Block} \\ \hline \end{array}$

   (A) A – IV, B – III, C – I, D – II (B) A – II, B – III, C – IV, D – I (C) A – III, B – IV, C – I, D – II (D) A – III, B – I, C – IV, D – II

   ›Reveal solutionSolution

   The problem asks to match each element (Hf, Ra, Am, At) to its block in the periodic table (s, p, d, f). The correct matches are: Hf (d-block), Ra (s-block), Am (f-block), At (p-block), so the answer is option (D).

   The periodic table is organized into blocks based on which subshell (s, p, d, or f) the element’s outermost electrons occupy. This is determined by the element’s electron configuration, which follows the Aufbau principle. The key is to know the position of each element in the periodic table, or to deduce its block from its group and period.

   Let’s match each element step by step:

   1. Hf (Hafnium)

      • Hafnium is element 72. It lies in period 6, group 4.
      • Group 4 elements are transition metals, which are always in the d-block.
      • Its electron configuration ends in $5d^2 6s^2$, confirming it’s a d-block element.
      • So A matches III (d-block).

   2. Ra (Radium)

      • Radium is element 88, in period 7, group 2.
      • Group 2 elements are alkaline earth metals, which are in the s-block (the last electron enters the s subshell).
      • Configuration: $[Rn] 7s^2$.
      • So B matches I (s-block).

   3. Am (Americium)

      • Americium is element 95, an actinide.
      • Actinides are part of the f-block (the 5f subshell is being filled).
      • Configuration: $[Rn] 5f^7 7s^2$.
      • So C matches IV (f-block).

   4. At (Astatine)

      • Astatine is element 85, in period 6, group 17 (halogens).
      • Group 17 elements are in the p-block (the last electron enters a p subshell).
      • Configuration: $[Xe] 4f^{14} 5d^{10} 6p^5$. …

   🔒 Start your 14-day free trial to see the full solution →
2. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.The energy of orbit X of Li$^{2+}$ (Z = 3) is $-2.18 \times 10^{-18}$ J. What is the radius of the same orbit (in $\text{\AA}$)? (A) 2.116 (B) 2.105 (C) 1.587 (D) 2.645

   ›Reveal solutionSolution

   The key idea is to use the Bohr model for hydrogen-like ions: energy depends on $n^2/Z^2$ and radius on $n^2/Z$. Given the energy of Li$^{2+}$, we first find the principal quantum number $n$, then compute the radius. The radius comes out to $1.587 \, \text{Å}$, so the correct option is (C).

   ---

   We are dealing with a hydrogen-like ion (Li$^{2+}$ has only one electron, so the Bohr model applies exactly). In the Bohr model, the energy of an electron in orbit $n$ for a nucleus of charge $Z$ is:

   $E_n = - \frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2}$

   But here the energy is given in joules. The ground-state energy of hydrogen ($Z=1$, $n=1$) is $-13.6 \, \text{eV} = -2.18 \times 10^{-18} \, \text{J}$. So we can write:

   $E_n = - \frac{Z^2}{n^2} \times (2.18 \times 10^{-18} \, \text{J})$

   For Li$^{2+}$, $Z = 3$, and we are told $E_n = -2.18 \times 10^{-18} \, \text{J}$. That means:

   $- \frac{3^2}{n^2} \times (2.18 \times 10^{-18}) = -2.18 \times 10^{-18}$

   Cancelling the common factor:

   $\frac{9}{n^2} = 1 \quad \Rightarrow \quad n^2 = 9 \quad \Rightarrow \quad n = 3$

   So the electron is in the third Bohr orbit.

   Now, the radius of the $n$th orbit for a hydrogen-like ion is:

   $r_n = \frac{n^2}{Z} \, a_0$

   where $a_0 = 0.529 \, \text{Å}$ is the Bohr radius.

   Plug in $n=3$, $Z=3$:

   $$$$

   🔒 Start your 14-day free trial to see the full solution →
3. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.The atomic numbers of four elements A, D, E, G are 4, 7, 8, 12 respectively. The decreasing order of electronegativity of these elements is (A) E, D, A, G (B) E, A, D, G (C) D, G, E, A (D) G, A, D, E

   ›Reveal solutionSolution

   Electronegativity increases across a period and decreases down a group. Using the periodic table positions of elements with atomic numbers 4 (Be), 7 (N), 8 (O), and 12 (Mg), the decreasing order is E, D, A, G, which corresponds to option (A).

   The key to ordering electronegativity is understanding the periodic trends. Electronegativity — the ability of an atom to attract shared electrons in a bond — increases as you move from left to right across a period (because nuclear charge increases while atomic radius decreases) and decreases as you move down a group (because the valence shell gets farther from the nucleus, reducing attraction). So the first step is to identify each element from its atomic number and place it in the periodic table.

   1. Identify the elements.

      Atomic number 4 is beryllium (Be, group 2, period 2).

      Atomic number 7 is nitrogen (N, group 15, period 2).

      Atomic number 8 is oxygen (O, group 16, period 2).

      Atomic number 12 is magnesium (Mg, group 2, period 3).

      So A = Be, D = N, E = O, G = Mg.

   2. Compare elements in the same period.

      Be, N, and O are all in period 2. Across a period, electronegativity increases from left to right. The order in period 2 (lowest to highest) is: Be < B < C < N < O < F. So among these three:

      Be (A) has the lowest, then N (D), then O (E) has the highest.

      So far: E > D > A.

   3. Place the element from the next period. …

   🔒 Start your 14-day free trial to see the full solution →
4. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.A metal was irradiated separately with radiation of wavelengths

   (a) 400 nm

   (b) 500 nm and

   (c) 600 nm respectively. Identify the wavelength(s) corresponding to which electrons are emitted from the surface of that metal (Work function of metal = 2.25 eV, $h = 6.6 \times 10^{-34}$ Js, $c = 3 \times 10^8$ ms$^{-1}$, $1$eV $= 1.6 \times 10^{-19}$ J) (A)

   (a) only (B)

   (c) only (C) (a),

   (b) &

   (c) (D)

   (a) &

   (b) only

   ›Reveal solutionSolution

   The photoelectric effect requires the incident photon energy to exceed the metal’s work function. Only wavelengths shorter than the threshold wavelength cause emission. Here, only 400 nm and 500 nm satisfy this, so the answer is (D).

   The core idea is the photoelectric effect: an electron is ejected from a metal surface only if the energy of an incident photon is at least equal to the work function $\phi$ of the metal. The photon energy is $E = \frac{hc}{\lambda}$, so for a given work function, there is a maximum wavelength (threshold wavelength $\lambda_0$) beyond which no emission occurs. Any wavelength shorter than $\lambda_0$ will cause emission; any longer will not.

   We are given $\phi = 2.25$ eV, $h = 6.6 \times 10^{-34}$ Js, $c = 3 \times 10^8$ m/s, and $1$ eV $= 1.6 \times 10^{-19}$ J. The three wavelengths are 400 nm, 500 nm, and 600 nm. We need to compare each photon’s energy to $\phi$.

   1. Find the threshold wavelength $\lambda_0$. Set the photon energy equal to the work function:

   $\frac{hc}{\lambda_0} = \phi$

   Convert $\phi$ to joules:

   $\phi = 2.25 \times 1.6 \times 10^{-19} = 3.6 \times 10^{-19} \text{ J}$

   Then

   $\lambda_0 = \frac{hc}{\phi} = \frac{(6.6 \times 10^{-34})(3 \times 10^8)}{3.6 \times 10^{-19}}$

   Calculate:

   $\lambda_0 = \frac{19.8 \times 10^{-26}}{3.6 \times 10^{-19}} = 5.5 \times 10^{-7} \text{ m} = 550 \text{ nm}$

   So the threshold wavelength is 550 nm. Any wavelength shorter than 550 nm will eject electrons; any longer will not.

   2. Compare each given wavelength to $\lambda_0$. …

   🔒 Start your 14-day free trial to see the full solution →
5. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.In an atomic spectrum of hydrogen, a series of lines with wavelengths at 656.46, 486.27, $x$ and 410.29 nm was obtained. What is the value of $x$ (in nm)? ($R_H = 1.097 \times 10^7 \, \mathrm{m^{-1}}$) (A) 453.15 (B) 449.32 (C) 434.17 (D) 428.37

   ›Reveal solutionSolution

   This tests the Rydberg formula for the Balmer series of hydrogen to find the missing wavelength ($n_i=5\to n_f=2$ transition) in a listed sequence of visible spectral lines. Answer: 434.17 nm.

   Concept and Intuition

   The Balmer series consists of hydrogen emission lines that end on the $n=2$ level, and these fall in the visible range — they are the most famous hydrogen lines (H-alpha, H-beta, H-gamma, H-delta at 656.3, 486.1, 434.1, 410.2 nm respectively, as commonly tabulated). The listed wavelengths 656.46, 486.27, $x$, 410.29 nm are exactly this sequence for $n_i=3,4,5,6$, so $x$ corresponds to the $n_i=5\to n_f=2$ transition (H-gamma).

   Step-by-Step Solution

   1. Rydberg formula: $\dfrac1\lambda=R_H\left(\dfrac1{n_f^2}-\dfrac1{n_i^2}\right)$.
   2. Recognize the series: with $n_f=2$, $n_i=3$ gives 656.46 nm, $n_i=4$ gives 486.27 nm, $n_i=6$ gives 410.29 nm — matching the given data confirms $n_f=2$ (Balmer) and that $x$ is the $n_i=5$ line.
   3. For $n_i=5$: $\dfrac1\lambda=R_H\left(\dfrac14-\dfrac1{25}\right)=R_H\left(\dfrac{25-4}{100}\right)=R_H\times\dfrac{21}{100}=0.21R_H$. …

   🔒 Start your 14-day free trial to see the full solution →
6. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.Wavelength of a photon emitted during electron transition from $n = 4$ state to $n = 2$ state in the hydrogen atom is $x$ nm. Wavelength of a photon emitted during electron transition from $n = 4$ state to $n = 1$ state in the same atom is $y$ nm. $\dfrac{y}{x}$ is equal to (A) 0.4 (B) 0.2 (C) 0.5 (D) 0.3

   ›Reveal solutionSolution

   This tests the Rydberg/Bohr formula for hydrogen-spectrum wavelengths; the ratio of the two transition wavelengths works out to 0.2.

   Concept and Intuition

   For the hydrogen atom, the energy released (and hence the wavenumber $1/\lambda$ of the emitted photon) during a transition from a higher level $n_2$ to a lower level $n_1$ is given by the Rydberg formula. A bigger energy jump means a shorter wavelength, so a transition all the way down to $n=1$ releases more energy (and gives a shorter wavelength) than one landing at $n=2$.

   Step-by-Step Solution

   1. Rydberg formula: $\dfrac{1}{\lambda}=R\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)$, with $n_1<n_2$.
   2. For $n=4\to n=2$ (wavelength $x$): $\dfrac{1}{x}=R\left(\dfrac{1}{2^2}-\dfrac{1}{4^2}\right)=R\left(\dfrac14-\dfrac1{16}\right)=R\cdot\dfrac{3}{16}$. …

   🔒 Start your 14-day free trial to see the full solution →
7. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.The energy of spectral line of lowest frequency in Lyman series of $\mathrm{Li^{2+}}$ spectrum is $x$ J. The energy of second spectral line in Balmer series of $\mathrm{He^{+}}$ spectrum is $y$ J. The ratio of $x$ and $y$ is (A) $3 : 1$ (B) $1 : 3$ (C) $1 : 9$ (D) $9 : 1$

   ›Reveal solutionSolution

   Both are hydrogen-like ions, so the Rydberg-type energy formula with $Z^2$ scaling applies; identifying the correct transitions (lowest-frequency Lyman line, second Balmer line) and computing gives $x:y=9:1$.

   Concept and Intuition

   For any hydrogen-like species (single electron, nuclear charge $Ze$), the energy released in a transition from level $n_2$ to $n_1$ ($n_2>n_1$) is

   $E = 13.6\,Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\ \text{eV}.$

   Within a series (fixed lower level $n_1$), the lowest-frequency (least energetic) line corresponds to the smallest jump, i.e. the transition from the level immediately above $n_1$. The Lyman series has $n_1=1$, so its lowest-frequency line is $2\to1$. The Balmer series has $n_1=2$; its lines in increasing energy (and frequency) order are $3\to2$ (first/weakest), $4\to2$ (second), $5\to2$ (third), etc. — so the second Balmer line is $4\to2$.

   Step-by-Step Solution

   1. Lyman, lowest-frequency line of $\mathrm{Li^{2+}}$ ($Z=3$): transition $2\to1$.

   $x = 13.6\times3^2\left(\frac1{1^2}-\frac1{2^2}\right) = 13.6\times9\times\frac34 = 13.6\times6.75 = 91.8\ \text{eV (in energy units, up to a common constant)}$

   2. Balmer, second line of $\mathrm{He^+}$ ($Z=2$): second line means $4\to2$ (first is $3\to2$). …

   🔒 Start your 14-day free trial to see the full solution →
8. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.Quantum number sets of four electrons I, II, III, IV are given below. The correct order of the energy of these electrons is I. $n = 3,\ l = 1,\ m_l = -1,\ m_s = +\frac{1}{2}$ II. $n = 4,\ l = 1,\ m_l = 0,\ m_s = +\frac{1}{2}$ III. $n = 4,\ l = 2,\ m_l = -2,\ m_s = +\frac{1}{2}$ IV. $n = 3,\ l = 2,\ m_l = -1,\ m_s = -\frac{1}{2}$ The correct answer is (A) I > IV > II > III (B) III > II > I > IV (C) III > IV > II > I (D) III > II > IV > I

   ›Reveal solutionSolution

   Applying the $(n+l)$ (Aufbau) rule with the n+l tie-breaker gives the energy order III > II > IV > I — testing whether the tie-break (lower $n$ wins for equal $n+l$) is applied correctly.

   Concept and Intuition

   In multi-electron atoms, orbital energy is not determined by $n$ alone (as in hydrogen) but follows the empirical $(n+l)$ rule: orbitals with a lower value of $n+l$ have lower energy. When two orbitals share the same $n+l$ value, the one with the smaller $n$ (and hence larger $l$) has lower energy — because a larger $l$ means the electron is, on average, farther from the nucleus in angular terms but the radial penetration effects work out so that lower-$n$/higher-$l$ combinations of equal $n+l$ sit lower in energy (e.g. $3d$ fills before $4p$... more precisely $4s$ before $3d$, but the classic comparison here is between $4p$ ($n+l=5$) and $3d$ ($n+l=5)$, where $3d$ is lower).

   Step-by-Step Solution

   1. Compute $n+l$ for each electron (the $m_l,m_s$ values don't affect orbital energy, only $n,l$ do):
      • I: $n=3,l=1$ (3p) $\Rightarrow n+l=4$
      • II: $n=4,l=1$ (4p) $\Rightarrow n+l=5$
      • III: $n=4,l=2$ (4d) $\Rightarrow n+l=6$
      • IV: $n=3,l=2$ (3d) $\Rightarrow n+l=5$
   2. Order by $n+l$ ascending (lower = lower energy): I(4) $<$ {II, IV}(5) $<$ III(6). …

   🔒 Start your 14-day free trial to see the full solution →
9. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.The wavelength of spectral line (X) of hydrogen spectrum is same as that of spectral line of $He^+$ spectrum corresponding to $n = 4 \to n = 2$ electron transition. The correct electron transition corresponding to X is (A) $n = 2 \to n = 1$ (B) $n = 3 \to n = 2$ (C) $n = 4 \to n = 2$ (D) $n = 3 \to n = 1$

   ›Reveal solutionSolution

   This tests the Rydberg formula scaled by $Z^2$ and matching two different hydrogenic spectra to the same photon energy/wavelength. The transition is $n=2\to n=1$.

   Concept and Intuition

   Every hydrogen-like ion (H, He$^+$, Li$^{2+}$, …) has energy levels $E_n = -\dfrac{13.6\,Z^2}{n^2}\ \text{eV}$. Two hydrogenic transitions emit the same wavelength exactly when their $Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$ values are equal — the $Z^2$ scaling is what lets a lower-$Z$ atom's low-lying transition mimic a higher-$Z$ ion's higher-lying one.

   Step-by-Step Solution

   1. For $He^+$ ($Z=2$), transition $n=4\to n=2$:

   $\frac{1}{\lambda_{He^+}} = R(2)^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right) = 4R\left(\frac14-\frac1{16}\right)=4R\cdot\frac{3}{16}=0.75R$

   2. For hydrogen ($Z=1$), we need a transition $n_1\to n_2$ with

   $\frac{1}{\lambda_H} = R(1)^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) = 0.75R$

   3. Test $n=2\to n=1$: $R\left(1-\frac14\right) = 0.75R$. This matches exactly. …

   🔒 Start your 14-day free trial to see the full solution →
10. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
    Q.If the energy required to remove an electron from the ground state of $He^+$ is x J, the energy (in J) required to remove an electron from the ground state of $Li^{2+}$ is (A) $\frac{3}{2}x$ (B) $\frac{2}{3}x$ (C) $\frac{9}{4}x$ (D) $\frac{4}{9}x$

    ›Reveal solutionSolution

    Both $He^+$ and $Li^{2+}$ are hydrogen-like (one electron); their ground-state ionization energies scale purely as $Z^2$, giving a factor of $9/4$. Answer: (C).

    Concept and Intuition

    $He^+$ ($Z=2$) and $Li^{2+}$ ($Z=3$) are both single-electron (hydrogen-like) species, so the Bohr-model ionization energy formula $E_n=13.6\,\dfrac{Z^2}{n^2}$ eV applies directly to each. For the ground state ($n=1$), the energy is simply proportional to $Z^2$.

    Step-by-Step Solution

    1. Ionization energy of $He^+$ ($Z=2$, $n=1$): $E_{He^+}=13.6\times2^2=13.6\times4=x$ (given).
    2. Ionization energy of $Li^{2+}$ ($Z=3$, $n=1$): $E_{Li^{2+}}=13.6\times3^2=13.6\times9$.
    3. Ratio: $\dfrac{E_{Li^{2+}}}{E_{He^+}}=\dfrac{13.6\times9}{13.6\times4}=\dfrac{9}{4}$.
    4. So $E_{Li^{2+}}=\dfrac{9}{4}x$. …

    🔒 Start your 14-day free trial to see the full solution →
11. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
    Q.Which of the following set of quantum numbers represent the electron with highest energy? (A) $n=3,\ l=0,\ m=0,\ s=+\frac{1}{2}$ (B) $n=3,\ l=1,\ m=1,\ s=-\frac{1}{2}$ (C) $n=3,\ l=2,\ m=1,\ s=+\frac{1}{2}$ (D) $n=4,\ l=0,\ m=0,\ s=-\frac{1}{2}$

    ›Reveal solutionSolution

    Comparing orbital energies from quantum numbers uses the $(n+l)$ rule (Aufbau/Madelung ordering): the set with the largest $(n+l)$ sum (and, on a tie, the larger $n$) is highest in energy.

    Concept and Intuition

    For multi-electron atoms, orbital energy isn't decided by $n$ alone — the $(n+l)$ rule says orbitals with a lower $(n+l)$ sum fill first (are lower in energy); when two orbitals share the same $(n+l)$, the one with smaller $n$ is lower. This is why 4s (n+l=4) fills before 3d (n+l=5), even though 3d has a smaller $n$.

    Step-by-Step Solution

    1. (A) $n=3,l=0$ (3s): $n+l=3$.
    2. (B) $n=3,l=1$ (3p): $n+l=4$.
    3. (C) $n=3,l=2$ (3d): $n+l=5$.
    4. (D) $n=4,l=0$ (4s): $n+l=4$. …

    🔒 Start your 14-day free trial to see the full solution →
12. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
    Q.Wavelength of a particular line in Balmer series of atomic spectrum of hydrogen is 656.4 nm. What is the wavelength (in nm) of corresponding line in the spectrum of $\mathrm{He^+}$? (A) 328.2 (B) 164.1 (C) 492.3 (D) 246.1

    ›Reveal solutionSolution

    The same Balmer transition in the hydrogen-like He$^+$ ion has wavelength scaled down by $Z^2=4$ from hydrogen's, giving $164.1\ nm$.

    Concept and Intuition

    The Rydberg formula for any hydrogen-like (single-electron) species is $\dfrac1\lambda = RZ^2\left(\dfrac1{n_1^2}-\dfrac1{n_2^2}\right)$. For the same pair of energy levels $n_1,n_2$ (the "corresponding line"), everything is identical between hydrogen and He$^+$ except the nuclear charge $Z$. Since $\dfrac1\lambda \propto Z^2$, the wavelength itself is inversely proportional to $Z^2$: a higher-charge nucleus pulls electrons in more tightly, so transition energies are larger and wavelengths shorter.

    Step-by-Step Solution

    1. For H ($Z=1$): $\dfrac{1}{\lambda_H} = R\left(\dfrac1{n_1^2}-\dfrac1{n_2^2}\right)$.
    2. For He$^+$ ($Z=2$): $\dfrac{1}{\lambda_{He^+}} = R(2)^2\left(\dfrac1{n_1^2}-\dfrac1{n_2^2}\right) = 4\times\dfrac{1}{\lambda_H}$. …

    🔒 Start your 14-day free trial to see the full solution →