JEE Physics Modern Physics MCQs Set 04

Answer: (b) 2948 Å

Explanation:
1. \( \lambda_0\propto\frac1\phi \).
2. \( \lambda_W = 5896\times\frac{2.53}{5.06} = 2948 \) Å.

Teacher's Note:
1. Threshold wavelength is inversely proportional to work function.
2. Double the work function halves the wavelength.

Answer: (a) the number of electrons emitted per second becomes one quarter of the initial number

Explanation:
1. Intensity falls as \( 1/r^{2} \), so the distance doubling reduces it to one quarter.
2. The photon energy (frequency) is unchanged, so each electron has the same maximum energy while the number of electrons per second drops to a quarter.

Teacher's Note:
1. Intensity controls the number of photoelectrons.
2. Frequency controls their energy.

Answer: (c) 1.45 μA

Explanation:
1. Photons per second \( = \frac{P\lambda}{hc} = \frac{4.5\times10^{-3}\times4\times10^{-7}}{1.99\times10^{-25}} = 9.06\times10^{15} \).
2. Electrons per second \( = 9.06\times10^{12} \), so \( I = 9.06\times10^{12}\times1.6\times10^{-19} = 1.45\ \mu A \).

Teacher's Note:
1. Count photons first.
2. Current = electrons per second × \( e \).

Answer: (c) 4 V

Explanation:
1. \( eV_0 = \tfrac12mv^2 \Rightarrow V_0 = \frac{v^2}{2(e/m)} \).
2. \( V_0 = \frac{1.44\times10^{12}}{2\times1.8\times10^{11}} = 4 V \).

Teacher's Note:
1. Stopping potential equals \( K.E \). per unit charge.
2. Use \( e/m \) directly.

Answer: (d) increasing the potential difference between the anode and the filament

Explanation:
1. Electrons gain \( \tfrac12mv^2 = eV \) in the accelerating field.
2. So a larger accelerating potential gives a higher speed; filament current only changes the number of electrons.

Teacher's Note:
1. Speed depends on the accelerating voltage.
2. Filament current controls emission rate.

Answer: (b) decrease 2 times

Explanation:
1. \( \lambda\propto\frac{1}{\sqrt V} \).
2. \( \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{10}{40}} = \frac12 \).

Teacher's Note:
1. \( \lambda = 12.27/\sqrt{V} \) Å.
2. Quadrupling \( V \) halves λ.

Answer: (c) A photon is a particle of zero momentum

Explanation:
1. A photon has momentum \( p = \frac h\lambda = \frac Ec \).
2. Its rest mass is zero, it moves at \( c \) in vacuum and its path bends in a gravitational field.

Teacher's Note:
1. Zero rest mass does not mean zero momentum.
2. Energy carries gravitational effect.

Answer: (a) \( 2.5\times10^{10} \)

Explanation:
1. Energy per photon \( = \frac{hc}{\lambda} = \frac{1.99\times10^{-25}}{5\times10^{-7}} = 4\times10^{-19} J \).
2. \( n = \frac{10^{-8}}{4\times10^{-19}} = 2.5\times10^{10} \).

Teacher's Note:
1. Divide power by photon energy.
2. hc = \( 1.99 \times 10^{-25} J m \).

Answer: (d) 3.9 V

Explanation:
1. \( K_{max} = 6-2.1 = 3.9 \) eV, so \( V_0 = 3.9 V \).

Teacher's Note:
1. eV₀ = hν − φ.
2. Stopping potential in volts equals \( K.E \). in eV.

Answer: (d) β-particle

Explanation:
1. \( \lambda = \frac{h}{mv} \) is largest for the smallest mass.
2. A β-particle (electron) has the smallest mass.

Teacher's Note:
1. \( \lambda ∝ 1/m \) at equal speed.
2. Compare masses.

Answer: (a) 6 V

Explanation:
1. \( \lambda = \frac{12.27}{\sqrt V} \) Å, so \( V = \left(\frac{12.27}{5}\right)^2 \approx 6 V \).

Teacher's Note:
1. Use λ(Å) = \( 12.27/\sqrt{V} \).
2. Wavelengths of 5 Å need low voltages.

Answer: (c) greater than the kinetic energy of the proton

Explanation:
1. Same λ means same momentum \( p \). \( K = \frac{p^2}{2m} \), so the lighter electron has the larger \( K \).

Teacher's Note:
1. \( K = p^{2}/2m \).
2. Equal momentum, unequal mass.

Answer: (a) momentum

Explanation:
1. Both have \( p = \frac h\lambda \).
2. Their energies and speeds differ.

Teacher's Note:
1. λ fixes momentum for both.
2. Energy of a photon is pc; of an electron \( p^{2}/2m \).

Answer: (d) \( \sqrt{\frac{2m_\alpha}{m_e}} \)

Explanation:
1. \( p = \sqrt{2mqV} \).
2. \( \frac{p_\alpha}{p_e} = \sqrt{\frac{m_\alpha(2e)}{m_e e}} = \sqrt{\frac{2m_\alpha}{m_e}} \).

Teacher's Note:
1. Kinetic energy gained is qV.
2. α has charge \( 2e \).

Answer: (b) 6

Explanation:
1. Photon energy \( = \frac{12400}{975} = 12.7 \) eV, which raises the atom to n = 4 (12.75 eV).
2. Lines \( = \frac{n(n-1)}{2} = 6 \).

Teacher's Note:
1. Find \( n \) from the photon energy.
2. Count all downward transitions.

Answer: (c) 11.1 eV

Explanation:
1. Allowed photon energies are differences of levels \( -13.6/n^{2} \): 0.66 eV (4→3), 1.89 eV (3→2), 10.2 eV (2→1), 12.09 eV (3→1), 13.6 eV (series limit).
2. 11.1 eV is not a difference of two levels.

Teacher's Note:
1. Compute the differences between levels.
2. The ionisation limit gives 13.6 eV.

Answer: (d) \( \frac1n,\ \frac1{n^2},\ n^2 \)

Explanation:
1. \( v\propto\frac1n \), \( |E|\propto\frac1{n^2} \), \( r\propto n^2 \).

Teacher's Note:
1. Remember \( v = 2.2\times 10^{6}/n m/s \).
2. Radius grows as \( n^{2} \).

Answer: (b) \( \frac5{27} \)

Explanation:
1. Lyman longest: \( \frac1\lambda = R\left(1-\frac14\right) = \frac{3R}4 \). Balmer longest: \( \frac1\lambda = R\left(\frac14-\frac19\right) = \frac{5R}{36} \).
2. Ratio \( = \frac{4/3R}{36/5R} = \frac{5}{27} \).

Teacher's Note:
1. Longest wavelength is the smallest energy jump in the series.
2. Use the Rydberg formula.

Answer: (d) \( 2\to1 \)

Explanation:
1. For He⁺ \( (Z = 2) \): \( \Delta E = 4\times13.6\left(\frac14-\frac1{16}\right) = 10.2 \) eV.
2. In \( H \) the 2 → 1 transition also gives 10.2 eV.

Teacher's Note:
1. Energy scales as \( Z^{2} \).
2. Compare energies, not shells.

Answer: (c) the energy difference between energy levels of hydrogen is much less

Explanation:
1. Characteristic X-rays arise from transitions to inner shells of heavy atoms with energy differences of several keV.
2. Hydrogen's level spacing is only a few eV.

Teacher's Note:
1. X-ray photons need keV energies.
2. Energy gaps grow roughly as \( Z^{2} \).

Answer: (c) the frequency is higher

Explanation:
1. Harder X-rays are more penetrating, which means higher photon energy.
2. Since \( E \) = hν, the frequency is higher and the wavelength shorter.

Teacher's Note:
1. Hardness depends on energy, not intensity.
2. \( E \) = hν.

Answer: (d) \( {}^{7}\mathrm{Li} \)

Explanation:
1. \( R\propto A^{1/3} \). For \( R/3 \) we need \( A = \frac{189}{27} = 7 \).
2. That is lithium-7.

Teacher's Note:
1. Volume is proportional to \( A \).
2. Cube root of the mass number ratio.

Answer: (b) 4.8

Explanation:
1. \( R = R_0A^{1/3} \Rightarrow R_{Cu} = 3.6\left(\frac{64}{27}\right)^{1/3} = 3.6\times\frac43 = 4.8 \) fm.

Teacher's Note:
1. Cube root of 64/27 is 4/3.
2. \( R ∝ A^{1/3} \).

Answer: (b) \( E_2>E_1 \)

Explanation:
1. Fission of a heavy nucleus releases energy, so the daughters are more tightly bound per nucleon.
2. Hence \( E_{2} \gt E_{1} \).

Teacher's Note:
1. Energy release requires an increase in binding energy.
2. Heavy nuclei lie on the falling part of the curve.

Answer: (a) \( M_2<2M_1 \)

Explanation:
1. Ca-40 has a higher binding energy per nucleon than Ne-20, so 2 × (mass of Ne-20) exceeds the mass of Ca-40.
2. Hence \( M_{2} \lt 2M_{1} \). Also M₁ < 10(mₙ + mₚ).

Teacher's Note:
1. Larger binding energy means a smaller mass.
2. Compare 2 Ne-20 with one Ca-40.