JEE Physics Atoms MCQs Set 06

Answer: (b) 5000 : 1

Explanation:
1. E = hc/λ, so E₁/E₂ = λ₂/λ₁ = 5000/1
2. Ratio = 5000 : 1.

Teacher's Note:
1. The shorter wavelength has the larger energy.
2. Do not square the ratio.

Answer: (b) the square of the atomic number of the element

Explanation:
1. Moseley's law: √ν = a(Z − b)
2. So ν ∝ (Z − b)², which varies approximately as the square of the atomic number.

Teacher's Note:
1. The square root of the frequency is linear in Z.
2. The law was used to arrange the elements by atomic number.

Answer: (a) X-rays have a wavelength of the order of the interatomic spacing

Explanation:
1. Diffraction by a crystal occurs only when the wavelength is comparable to the spacing between atomic planes (about 1 Å)
2. X-ray wavelengths are of this order.

Teacher's Note:
1. Nuclear size is about 10⁻¹⁵ m, far smaller than X-ray wavelengths.
2. Penetrating power alone does not give diffraction.

Answer: (d) the potential difference between cathode and target

Explanation:
1. Hardness depends on photon energy
2. Photon energy is controlled by the accelerating potential difference between cathode and target.

Teacher's Note:
1. Filament current controls intensity.
2. The tube is evacuated so that the pressure is very low.

Answer: (d) γ-rays

Explanation:
1. γ-rays have the highest photon energies and the least ionising power, so they penetrate the most
2. α-particles are stopped by paper and β-particles by thin metal sheets.

Teacher's Note:
1. Order of penetration: α < β < X-rays < γ.
2. Highly ionising radiation loses energy quickly.

Answer: (d) intensity

Explanation:
1. More electrons striking the target produce more photons
2. The intensity increases while the photon energy, frequency and wavelength stay the same.

Teacher's Note:
1. Penetration, frequency and wavelength depend on voltage.
2. Intensity depends on the number of electrons per second.

Answer: (c) n = 3 to n = 1

Explanation:
1. K_β is emitted when an electron falls from the M shell (n = 3) into the K shell (n = 1).

Teacher's Note:
1. K_α is n = 2 to n = 1 and K_β is n = 3 to n = 1.
2. K_β photons have more energy than K_α photons.

Answer: (c) will be undeviated

Explanation:
1. X-ray photons are neutral, so no magnetic force acts on them and they go straight.

Teacher's Note:
1. A magnetic field acts only on moving charges.
2. The same is true of electric fields.

Answer: (a) 0.25 Å

Explanation:
1. λmin = 12400/40000 = 0.31 Å
2. No X-ray can have a wavelength shorter than 0.31 Å
3. 0.25 Å is shorter than the cut-off, so it is not possible.

Teacher's Note:
1. All wavelengths longer than λmin are possible.
2. Compute the cut-off first.

Answer: (b) 3 × 10⁸ m/s

Explanation:
1. X-rays are electromagnetic waves and move with the speed of light, whatever the voltage.

Teacher's Note:
1. Voltage changes the photon energy, not the speed.
2. Speed of light = 3 × 10⁸ m/s.

Answer: (d) not deflect

Explanation:
1. X-rays carry no charge, so the magnetic field exerts no force on them.

Teacher's Note:
1. Only charged particles feel the Lorentz force.
2. Photons travel in straight lines in fields.

Answer: (d) 5 kV

Explanation:
1. V = 12400/λmin(Å) = 12400/2.5 = 4960 V
2. This is about 5 kV.

Teacher's Note:
1. 2.5 × 10⁻¹⁰ m is 2.5 Å.
2. A smaller wavelength would need a larger voltage.

Answer: (b) the target material

Explanation:
1. The energy levels, and so the line wavelengths, depend on the atomic number of the target.

Teacher's Note:
1. Different target elements give different line wavelengths.
2. This is why the lines are called characteristic.

Answer: (b) 2.48 × 10⁻¹⁰ m

Explanation:
1. λmin = 12400/5000 = 2.48 Å = 2.48 × 10⁻¹⁰ m.

Teacher's Note:
1. 1 Å = 10⁻¹⁰ m.
2. Use hc/e = 12400 V Å.

Answer: (a) 0.62 Å

Explanation:
1. λmin = 12400/20000 = 0.62 Å.

Teacher's Note:
1. Use λmin(Å) = 12400/V(volt).
2. Doubling the voltage halves the wavelength.

Answer: (b) \(\frac{hc}{eV}\)

Explanation:
1. At the cut-off the whole kinetic energy eV becomes one photon: eV = hc/λmin
2. λmin = hc/eV.

Teacher's Note:
1. Dimensionally hc/(energy) is a length.
2. Both e and V must appear in the denominator.

Answer: (c) increase the voltage applied between cathode and anticathode

Explanation:
1. Hardness depends on the energy of electrons striking the target
2. This is raised by raising the voltage between the cathode and the anode.

Teacher's Note:
1. Filament voltage and current change the intensity.
2. Harder means more penetrating.

Answer: (a) 1 Å

Explanation:
1. Diffraction needs the obstacle spacing to be comparable to the wavelength
2. X-ray wavelengths are of order 1 Å.

Teacher's Note:
1. Crystal spacings are about 1 to 3 Å.
2. Visible light is far too long for crystals.

Answer: (c) 1 Å

Explanation:
1. X-ray wavelengths range from about 0.1 Å to 100 Å
2. 1 Å is within this range; 10⁴ Å is infrared and 10⁻⁴ Å is a γ-ray.

Teacher's Note:
1. Visible light is 4000 to 7000 Å.
2. X-rays are about 10⁻¹⁰ m.

Answer: (b) 3 × 10⁻⁸ cm

Explanation:
1. Bragg's law: 2d sin θ = nλ
2. λ = 2 × 3 × 10⁻⁸ × sin 30° / 1 = 2 × 3 × 10⁻⁸ × 0.5 = 3 × 10⁻⁸ cm.

Teacher's Note:
1. The glancing angle is measured from the plane, not the normal.
2. sin 30° = 1/2.

Answer: (b) the energy of the electron

Explanation:
1. λmin = hc/E, where E is the kinetic energy of the incident electron
2. The target's atomic number does not enter.

Teacher's Note:
1. Atomic number decides the characteristic lines only.
2. Energy of electron = eV.

Answer: (b) keV

Explanation:
1. X-ray photons of wavelength 1 Å have energy 12.4 keV
2. So the order is keV.

Teacher's Note:
1. Visible light photons are a few eV.
2. γ-ray photons are of the order of MeV.

Answer: (a) 1.93 Å

Explanation:
1. λmin = 12400/6400 = 1.94 Å, i.e. about 1.93 Å.

Teacher's Note:
1. Use λ = 12400/V in Å.
2. Lower voltage gives a longer wavelength.