JEE Physics Atoms MCQs Set 09

Answer: (b) 1.51 eV

Explanation:
1. Bohr quantisation: \( L=\frac{nh}{2\pi} \), so n = 3.
2. Kinetic energy in the nth orbit of hydrogen \( =\frac{13.6}{n^2}=\frac{13.6}{9}=1.51\ \text{eV} \).

Teacher's Note:
1. Angular momentum fixes the orbit number n.
2. In Bohr's model, kinetic energy is the negative of the total energy, 13.6/n² eV for hydrogen.

Answer: (c) 2.42 × 10⁻¹⁹

Explanation:
1. Energy needed from level n is \( \frac{2.18\times10^{-18}}{n^2} \).
2. For n = 3: \( \frac{2.18\times10^{-18}}{9}=2.42\times10^{-19}\ \text{J} \).

Teacher's Note:
1. Binding energy in level n falls as 1/n².
2. The excited electron needs only one-ninth of the ground state ionisation energy.

Answer: (d) 4

Explanation:
1. Bohr radius \( r_n=n^2a_0 \).
2. \( (n+1)^2-n^2=(n-1)^2\Rightarrow 2n+1=n^2-2n+1 \).
3. \( n^2-4n=0\Rightarrow n=4 \) (n = 0 is not allowed).

Teacher's Note:
1. Radii grow as the square of the quantum number.
2. Write the condition as an equation in n and solve the quadratic.

Answer: (c) n² a0

Explanation:
1. From quantisation and force balance, \( r_n=\frac{n^2h^2\varepsilon_0}{\pi m e^2}=n^2a_0 \).

Teacher's Note:
1. Radius is proportional to n² and inversely to the nuclear charge Z.
2. For hydrogen Z = 1, so r_n = n² a0.

Answer: (a) 3.4 eV

Explanation:
1. \( E_n=-\frac{13.6}{n^2}\text{ eV} \).
2. For n = 2 the energy is -3.4 eV, so 3.4 eV must be supplied to free the electron.

Teacher's Note:
1. Ionisation energy from level n is 13.6/n² eV.
2. Do not confuse it with the ground state value.

Answer: (a) 63.75 eV

Explanation:
1. Excitation energies to the first excited state: H needs 10.2 eV; He+ (Z = 2) needs \( 4\times10.2=40.8\ \text{eV} \). Total required = 51 eV.
2. The minimum K corresponds to a perfectly inelastic collision, in which the energy lost is the fraction \( \frac{m_2}{m_1+m_2}=\frac45 \) of K.
3. \( \frac45K=51\Rightarrow K=63.75\ \text{eV} \).

Teacher's Note:
1. Maximum loss of kinetic energy occurs when the two bodies move together after the collision.
2. Loss = (reduced mass / m1) x K = 4/5 K here.

Answer: (b) 12.1 eV

Explanation:
1. \( \Delta E=13.6\left(1-\frac19\right)=12.09\approx12.1\ \text{eV} \).

Teacher's Note:
1. The atom can only absorb an amount of energy equal to the gap between two levels.
2. Here the gap is between n = 1 and n = 3.

Answer: (c) 10.2 eV of the incident electron energy is absorbed by the H atom and the electron comes out with 1.0 eV

Explanation:
1. The atom can absorb only 10.2 eV (n = 1 to 2); the next level needs 12.09 eV, which is more than 11.2 eV.
2. So 10.2 eV is absorbed and the electron carries away \( 11.2-10.2=1.0\ \text{eV} \).

Teacher's Note:
1. The atom accepts only discrete energy amounts.
2. The remainder of the energy stays with the electron.

Answer: (b) 3.3 m/s

Explanation:
1. Photon energy \( =10.2\ \text{eV}=1.63\times10^{-18}\ \text{J} \).
2. Photon momentum \( p=\frac Ec=\frac{1.63\times10^{-18}}{3\times10^8}=5.4\times10^{-27}\ \text{kg m/s} \).
3. By momentum conservation, \( v=\frac{p}{m}=\frac{5.4\times10^{-27}}{1.67\times10^{-27}}\approx3.3\ \text{m/s} \).

Teacher's Note:
1. The atom recoils with the momentum of the emitted photon.
2. Photon momentum is E/c.

Answer: (b) 20.4 eV

Explanation:
1. For equal masses, the largest energy that can be lost in a collision is half of the initial kinetic energy.
2. The atom must receive at least 10.2 eV to reach n = 2.
3. \( \frac12K=10.2\Rightarrow K=20.4\ \text{eV} \).

Teacher's Note:
1. Equal masses share the kinetic energy: at most half is transferable.
2. The threshold excitation energy of hydrogen is 10.2 eV.

Answer: (b) 4.2 m/s

Explanation:
1. 5th excited state is n = 6. Photon energy \( =13.6\left(1-\frac1{36}\right)=13.22\ \text{eV}=2.12\times10^{-18}\ \text{J} \).
2. Momentum \( =\frac Ec=7.06\times10^{-27} \), so \( v=\frac{7.06\times10^{-27}}{1.67\times10^{-27}}\approx4.2\ \text{m/s} \).

Teacher's Note:
1. Remember that the nth excited state is level n + 1.
2. Recoil speed is p/m.

Answer: (c) 1 Å

Explanation:
1. X-rays have wavelengths roughly from 0.1 Å to a few hundred Å.
2. 10,000 Å is infrared/visible and 1000 Å is ultraviolet; 10⁻² Å is in the gamma-ray range; 1 Å lies in the X-ray band.

Teacher's Note:
1. Order the electromagnetic spectrum by wavelength to place each option.
2. X-rays sit between ultraviolet and gamma rays.

Answer: (d) increase in their frequency

Explanation:
1. All X-rays travel at the speed of light, so velocity cannot change.
2. Higher-frequency (shorter-wavelength) X-rays are 'harder' and penetrate matter more.

Teacher's Note:
1. Intensity governs how many photons there are, not how deeply each penetrates.
2. Higher potential across the tube gives a higher maximum frequency and harder X-rays.

Answer: (c) νKβ = νKα + νLα

Explanation:
1. Kα arises from the L to K transition, Kβ from M to K, and Lα from M to L.
2. \( E_{M\to K}=E_{L\to K}+E_{M\to L} \), hence \( \nu_{K\beta}=\nu_{K\alpha}+\nu_{L\alpha} \).

Teacher's Note:
1. Energy differences add around a cycle of levels K, L, M.
2. The largest energy gap (Kβ) is the sum of the two smaller ones.

Answer: (c) 3.2 × 10⁻³² kg

Explanation:
1. Maximum photon energy \( =eV=18\ \text{keV}=18\times10^3\times1.6\times10^{-19}=2.88\times10^{-15}\ \text{J} \).
2. Equivalent mass \( m=\frac{E}{c^2}=\frac{2.88\times10^{-15}}{9\times10^{16}}=3.2\times10^{-32}\ \text{kg} \).

Teacher's Note:
1. The most energetic photon gets the whole kinetic energy of one electron.
2. Mass equivalent is E/c².

Answer: (d) doubly ionised lithium

Explanation:
1. Transition energy for a hydrogen-like species: \( \Delta E=13.6Z^2\left(1-\frac14\right)\text{ eV} \).
2. Z = 3 for doubly ionised lithium is the largest, giving the largest energy and the shortest wavelength.

Teacher's Note:
1. Energy of the line scales as Z² (the small reduced-mass differences between H and D do not matter).
2. Shortest wavelength means largest photon energy.

Answer: (c) 1 : 8

Explanation:
1. For Bohr orbits \( r\propto n^2 \), so radii R and 4R correspond to n = 1 and n = 2.
2. Time period \( T=\frac{2\pi r}{v}\propto n^3 \).
3. Ratio of periods = 1 : 8.

Teacher's Note:
1. Period in a Coulomb orbit follows Kepler's law T² proportional to r³.
2. With r in the ratio 1 : 4, T goes as (1 : 4)^(3/2) = 1 : 8.

Answer: (b) 1 + 2 + 3 + ... + n

Explanation:
1. The nth excited state is the level with quantum number n + 1.
2. The number of lines from level N is \( \frac{N(N-1)}{2}=\frac{(n+1)n}{2} \).
3. This equals the sum 1 + 2 + ... + n.

Teacher's Note:
1. Count possible downward transitions: N - 1 from the top level, N - 2 from the next, and so on.
2. The sum of the first n natural numbers is n(n + 1)/2.

Answer: (b) x = 9/4

Explanation:
1. First jump 3 to 2: \( E_1=13.6\left(\frac14-\frac19\right)=\frac{5}{36}(13.6) \). Second jump 2 to 1: \( E_2=\frac34(13.6) \).
2. Energy ratio \( z=\frac{E_1}{E_2}=\frac{5/36}{3/4}=\frac{5}{27} \). Momentum ratio y = z = 5/27, since p = E/c.
3. Wavelength ratio \( x=\frac{\lambda_1}{\lambda_2}=\frac{E_2}{E_1}=\frac{27}{5} \), which is not 9/4, so (b) is wrong.

Teacher's Note:
1. Wavelength is inversely proportional to photon energy, while momentum and energy are directly proportional.
2. Check each statement against 5/27 and 27/5.

Answer: (b) The total energy of the shell decreases with an increase in the value of n

Explanation:
1. With the usual zero at infinity, \( E_n=-\frac{13.6}{n^2} \) and \( U_1=-27.2\ \text{eV} \).
2. Moving the zero of potential energy to the ground state raises every energy by 27.2 eV, so \( E_n=27.2-\frac{13.6}{n^2} \).
3. Then E increases with n (A true), differences are unchanged (C true), and E1 = 27.2 - 13.6 = 13.6 eV (D true). Statement (B) is incorrect.

Teacher's Note:
1. Changing the reference level shifts all energies by the same constant.
2. Energy differences are independent of the reference level.

Answer: (c) 2.05 eV

Explanation:
1. Photon momentum \( p=\frac{h}{\lambda}=\frac{6.63\times10^{-34}}{10^{-11}}=6.63\times10^{-23}\ \text{kg m/s} \).
2. Recoil kinetic energy \( K=\frac{p^2}{2m}=\frac{(6.63\times10^{-23})^2}{2\times4\times1.66\times10^{-27}}=3.3\times10^{-19}\ \text{J} \).
3. \( K\approx\frac{3.3\times10^{-19}}{1.6\times10^{-19}}\approx2.05\ \text{eV} \).

Teacher's Note:
1. The atom recoils with the momentum of the photon, so its kinetic energy is p²/2m.
2. 1 u = 1.66 x 10^-27 kg.

Answer: (a) 1 : 10

Explanation:
1. Electron wavelength \( \lambda_e=\frac{12.27}{\sqrt{10^4}}=0.123\ \text{Å} \).
2. Shortest X-ray wavelength \( \lambda_{min}=\frac{hc}{eV}=\frac{12400}{10^4}=1.24\ \text{Å} \).
3. Ratio \( \lambda_e:\lambda_{min}=0.123:1.24\approx1:10 \).

Teacher's Note:
1. The cut-off wavelength is set by the electron's whole kinetic energy converted into one photon.
2. The de Broglie wavelength of the electron comes from p = sqrt(2meV).

Answer: (c) the difference between the wavelength of Kα and λmin will increase

Explanation:
1. Kα wavelength depends only on the anode material, so it does not change.
2. \( \lambda_{min}=\frac{hc}{eV} \) decreases when V increases.
3. Hence the difference \( \lambda_{K\alpha}-\lambda_{min} \) increases.

Teacher's Note:
1. Characteristic lines are fixed by the atomic structure of the target, not by the tube voltage.
2. Cut-off wavelength is inversely proportional to the voltage.

Answer: (c) 1.52 Å

Explanation:
1. Moseley's law: \( \frac1\lambda\propto(Z-1)^2 \) for Kα lines.
2. \( \frac{\lambda_{Cu}}{\lambda_{Mo}}=\left(\frac{42-1}{29-1}\right)^2=\left(\frac{41}{28}\right)^2=2.14 \).
3. \( \lambda_{Cu}=0.71\times2.14=1.52\ \text{Å} \).

Teacher's Note:
1. For Kα lines the effective nuclear charge is (Z - 1) because of screening by the remaining K electron.
2. Lighter elements emit longer wavelengths.

Answer: (b) electrons accelerate through a potential difference of 18.75 kV in the Coolidge tube

Explanation:
1. \( V=\frac{hc}{e\lambda_{min}}=\frac{1240\ \text{eV nm}}{0.0663\ \text{nm}}=18.7\ \text{kV} \), so (b) is correct and (a) is wrong.
2. The electrons' de Broglie wavelength \( =\frac{12.27}{\sqrt{18700}}\ \text{Å}=0.09\ \text{Å} \), which is neither 10 μm nor 0.01 Å.

Teacher's Note:
1. The cut-off wavelength gives the accelerating voltage directly, V = 12.4 / λmin (kV with λ in Å).
2. Check the de Broglie wavelength to rule out (c) and (d).