Bohr Model and Hydrogen Spectra — NEET Physics PYQ MCQs with Solutions

Solution: $E_n = \frac{-Rhc}{n^2} = \frac{-13.6}{n^2}$ eV (since $Rhc = 13.6$ eV)

Solution: The Bohr model is valid only for hydrogen and single-electron (hydrogen-like) species such as He⁺ and deuterium. Ne⁺ has many electrons, so the model is not valid for it.

Solution: Using the reduced mass of the muon–proton system: $m' = \frac{207m_e \times 1836m_e}{207m_e + 1836m_e} \approx 186m_e$ Radius $\propto \frac{1}{m}$: $r_1' = \frac{0.51\text{ Å}}{186} \approx 2.56 \times 10^{-13}$ m Energy $\propto m$: $E_1' = 186 \times (-13.6\text{ eV}) \approx -2.5$ keV — closest to $-2.8$ keV.

Solution: In a Bohr orbit: $KE = -E$ and $PE = 2E$ With $E = -3.4$ eV: $KE = 3.4$ eV and $PE = -6.8$ eV

Solution: $KE_n = \frac{Rhc}{n^2}$ and $TE_n = \frac{-Rhc}{n^2}$ $KE_n : TE_n = 1 : -1$

Solution: Last line of Balmer series ($n_1 = 2, n_2 = \infty$): $\frac{1}{\lambda_B} = \frac{R}{4}$ Last line of Lyman series ($n_1 = 1, n_2 = \infty$): $\frac{1}{\lambda_L} = R$ $\frac{\lambda_B}{\lambda_L} = 4$

Solution: $\frac{1}{\lambda} = R\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = \frac{5R}{36}$ $\frac{1}{\lambda'} = R\left(\frac{1}{3^2} - \frac{1}{4^2}\right) = \frac{7R}{144}$ $\lambda' = \frac{5}{36} \times \frac{144}{7}\lambda = \frac{20}{7}\lambda$

Solution: Radius in a magnetic field: $R = \frac{\sqrt{2mE}}{qB}$, so $E = \frac{q^2B^2R^2}{2m}$ For the $\alpha$-particle ($q = 2e$, $m = 4m_p$): $\frac{E_\alpha}{E_p} = \frac{(2e)^2}{4m_p} \cdot \frac{m_p}{e^2} = 1$ $E_\alpha = 1$ MeV

Solution: Longest Lyman line (2→1): $\frac{1}{\lambda_1} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}$ Longest Balmer line (3→2): $\frac{1}{\lambda_2} = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36}$ $\frac{\lambda_1}{\lambda_2} = \frac{5}{36} \times \frac{4}{3}... = \frac{5}{27}$

Solution: Speed of electron in nth orbit: $v = \frac{2.18 \times 10^6 \, Z}{n}$ m/s $v = \frac{2.18 \times 10^6 \times 2}{3} \approx 1.46 \times 10^6$ m/s

Solution: Photon energy $= \frac{12375}{975} \approx 12.75$ eV — this excites the electron from $n = 1$ to $n = 4$. Number of spectral lines $= \frac{n(n-1)}{2} = \frac{4 \times 3}{2} = 6$

Solution: $\lambda_L = \frac{1}{R(1 - 1/4)} = \frac{4}{3R}$; $\lambda_B = \frac{1}{R(1/4 - 1/9)} = \frac{36}{5R}$ $\frac{\lambda_L}{\lambda_B} = \frac{4}{3R} \times \frac{5R}{36} = \frac{5}{27}$

Solution: Energy released in the 2→1 transition: $E = -3.4 - (-13.6) = 10.2$ eV $h\nu_0 = E - eV_0 = 10.2 - 3.57 = 6.63$ eV $\nu_0 = \frac{6.63 \times 1.6 \times 10^{-19}}{6.67 \times 10^{-34}} = 1.6 \times 10^{15}$ Hz

Solution: For $\lambda_1$ (4→3): $\frac{hc}{\lambda_1} = 13.6\left(\frac{1}{9} - \frac{1}{16}\right) = 13.6 \times \frac{7}{144}$ For $\lambda_2$ (3→2): $\frac{hc}{\lambda_2} = 13.6\left(\frac{1}{4} - \frac{1}{9}\right) = 13.6 \times \frac{5}{36}$ $\frac{\lambda_1}{\lambda_2} = \frac{5}{36} \times \frac{144}{7} = \frac{20}{7}$

Solution: Photon energy: $\frac{hc}{\lambda} = Rhc\left(1 - \frac{1}{25}\right) = \frac{24Rhc}{25}$, so $\frac{1}{\lambda} = \frac{24R}{25}$ By momentum conservation, the atom recoils with $v = \frac{h}{m\lambda} = \frac{24hR}{25m}$

Solution: H first Lyman line: $\frac{1}{\lambda} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}$ H-like ion, second Balmer line (4→2): $\frac{1}{\lambda} = Z^2R\left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3Z^2R}{16}$ Equating: $\frac{3}{4} = \frac{3Z^2}{16} \Rightarrow Z = 2$

Solution: $E = \frac{-13.6\,Z^2}{n^2}$ with $Z = 2$, $n = 2$: $E = \frac{-13.6 \times 4}{4} = -13.6$ eV

Solution: $\frac{n(n-1)}{2} = 6 \Rightarrow n = 4$ Maximum wavelength corresponds to the minimum energy difference — the transition from $n = 4$ to $n = 3$.

Solution: $E_1 = -13.6$ eV; $E_2 = -3.4$ eV Excitation energy $= E_2 - E_1 = -3.4 + 13.6 = 10.2$ eV

Solution: For $n = 2$: $E_2 = \frac{-13.6}{4} = -3.4$ eV $KE = -E = 3.4$ eV

Solution: Free electrons liberated from the cathode are accelerated by the applied potential and collide with neutral gas atoms/molecules, ionising them.

Solution: $E_2 = -13.6 + 12.1 = -1.5$ eV $= \frac{-13.6}{n^2} \Rightarrow n^2 \approx 9 \Rightarrow n = 3$ Spectral lines $= \frac{n(n-1)}{2} = \frac{3 \times 2}{2} = 3$

Solution: In any Bohr orbit, the kinetic energy equals the negative of the total energy: $KE = -E = -(-3.4) = 3.4$ eV

Solution: $\Delta E = E_3 - E_2 = -\frac{13.6}{9} - \left(-\frac{13.6}{4}\right) = -1.51 + 3.4 \approx 1.9$ eV

Solution: Bohr postulated that electrons revolve only in orbits where the angular momentum is an integral multiple of $\frac{h}{2\pi}$ — the quantisation of angular momentum.

Solution: Photons are emitted only in downward transitions, ruling out (a) and (c). $\Delta E_{2\to1} = 13.6 \times \frac{3}{4} = 10.2$ eV; $\Delta E_{6\to2} = 13.6 \times \frac{2}{9} \approx 3.02$ eV The 2→1 transition gives the maximum-energy photon.

Solution: Balmer series ends at $n = 2$: the first line is 3→2, the second line is 4→2. Emission (not absorption) occurs in a downward jump.

Solution: Coulomb force provides the centripetal force: $\frac{mv^2}{a_0} = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{a_0^2}$ $v = \frac{e}{\sqrt{4\pi\varepsilon_0 a_0 m}}$

Solution: $E_2 = \frac{-13.6}{2^2} = -3.4$ eV

Solution: $r_n \propto n^2$. For the first excited state $n = 2$: $r_2 = 4r_1$