JEE Physics Dual Nature Of Radiation and Matter MCQs Set 09

Answer: (b) 2.5 × 10¹¹

Explanation:
1. Energy of one photon is \( \dfrac{hc}{\lambda} = \dfrac{6.63\times10^{-34}\times3\times10^{8}}{5\times10^{-7}} = 3.98\times10^{-19}\ \text{J} \).
2. Number of photons \( n = \dfrac{10^{-7}}{3.98\times10^{-19}} = 2.5\times10^{11} \).

Teacher's Note:
1. Total energy divided by the energy of one photon gives the number of photons.
2. Convert angstrom to metres before substituting.

Answer: (b) 1.6 × 10⁻¹¹ m

Explanation:
1. \( \lambda = \dfrac{h}{\sqrt{2mK}} \) with \( K = 3\times1.6\times10^{-19} = 4.8\times10^{-19}\ \text{J} \).
2. \( \sqrt{2\times1.7\times10^{-27}\times4.8\times10^{-19}} = 4.04\times10^{-23} \).
3. \( \lambda = \dfrac{6.6\times10^{-34}}{4.04\times10^{-23}} = 1.6\times10^{-11}\ \text{m} \).

Teacher's Note:
1. Convert electron-volts to joules before using the formula.
2. Heavier particles of the same energy have shorter wavelengths.

Answer: (b) intensity of light

Explanation:
1. Each photon ejects at most one electron, and the number of photons per second is proportional to the intensity.
2. So the number of photoelectrons emitted per second is proportional to the intensity.
3. The frequency decides only their maximum kinetic energy.

Teacher's Note:
1. Intensity controls the number of photoelectrons; frequency controls their energy.
2. Below the threshold frequency no electrons are emitted at all.

Answer: (a) 1.53 Å

Explanation:
1. \( \lambda = \dfrac{h}{\sqrt{2meV}} = \dfrac{6.623\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times1.6\times10^{-19}\times64}} \).
2. The denominator is \( \sqrt{1.864\times10^{-47}} = 4.32\times10^{-24} \).
3. \( \lambda = 1.53\times10^{-10}\ \text{m} = 1.53\ \text{Å} \).

Teacher's Note:
1. A quick check is \( \lambda = 12.27/\sqrt{V}\ \text{Å} = 12.27/8 \approx 1.53\ \text{Å} \).
2. Wavelength falls as the accelerating voltage rises.

Answer: (a) zero

Explanation:
1. The maximum kinetic energy is \( 6-4 = 2\ \text{eV} \) and it belongs to electrons from the surface.
2. Electrons that lose energy by collisions inside the metal before escaping come out with less energy, and some leave with almost none.
3. Hence the minimum kinetic energy is zero.

Teacher's Note:
1. Einstein's equation gives the maximum kinetic energy only.
2. Do not confuse the maximum with the minimum.

Answer: (b) 2N and T

Explanation:
1. Doubling the intensity doubles the number of photons per second, so the number of photoelectrons doubles to 2N.
2. The maximum kinetic energy \( K = h\nu-\phi \) depends only on frequency, so it stays T.

Teacher's Note:
1. Intensity affects the photocurrent; frequency affects the kinetic energy.
2. The two are independent in the photoelectric effect.

Answer: (b) 2E/c

Explanation:
1. The incident radiation carries momentum \( p = E/c \).
2. For total reflection the momentum reverses, so the change in momentum is \( \dfrac{E}{c} - \left(-\dfrac{E}{c}\right) = \dfrac{2E}{c} \).

Teacher's Note:
1. A perfectly absorbing surface receives E/c and a perfect reflector receives twice as much.
2. This is the basis of radiation pressure.

Answer: (a) ultraviolet region

Explanation:
1. Energy of the incident photon \( = \phi + eV_0 = 6.2 + 5 = 11.2\ \text{eV} \).
2. Wavelength \( \lambda = \dfrac{12400}{11.2}\ \text{Å} \approx 1100\ \text{Å} \).
3. This is far below 4000 Å and well above the X-ray range, so it is ultraviolet.

Teacher's Note:
1. Use \( \lambda(\text{Å}) = 12400/E(\text{eV}) \) to locate the region quickly.
2. Visible light corresponds to about 1.6 to 3.1 eV.

Answer: (c) ν₁ < ν₂ < ν₃

Explanation:
1. The stopping (retarding) potential satisfies \( eV_0 = h\nu-\phi \), so a higher frequency needs a larger retarding potential.
2. In the graph, the curve for ν₃ cuts the axis at the largest retarding potential, then ν₂, then ν₁.
3. Hence \( \nu_1 < \nu_2 < \nu_3 \).

Teacher's Note:
1. The farther left the curve meets the axis, the larger the stopping potential.
2. Saturation current is the same because the intensity is the same.

Answer: (d) V/8 volt

Explanation:
1. \( \lambda = \dfrac{h}{\sqrt{2mqV}} \), so equal wavelengths need equal \( mqV \).
2. For the proton mq = m·e. For the α-particle \( m_\alpha q_\alpha = 4m\times2e = 8me \).
3. So \( V_\alpha = \dfrac{V}{8} \).

Teacher's Note:
1. Wavelength depends on the product of mass, charge and accelerating voltage.
2. Heavier and more charged particles need a smaller voltage for the same λ.

Answer: (d)

Explanation:
1. The de Broglie relation is \( p = \dfrac{h}{\lambda} \).
2. So p is inversely proportional to λ, giving a rectangular hyperbola that falls as λ increases.
3. That is the curve in option (d).

Teacher's Note:
1. Inverse proportionality gives a hyperbola and never touches either axis.
2. A straight line would mean p is linear in λ.

Answer: (c) K₁ < K₂/2

Explanation:
1. \( K_1 = \dfrac{hc}{\lambda_1}-\phi = \dfrac{hc}{2\lambda_2}-\phi \) and \( K_2 = \dfrac{hc}{\lambda_2}-\phi \).
2. Then \( 2K_1 = \dfrac{hc}{\lambda_2}-2\phi = K_2-\phi \).
3. Since \( \phi > 0 \), \( 2K_1 < K_2 \), that is \( K_1 < K_2/2 \).

Teacher's Note:
1. Kinetic energy is not proportional to frequency because of the work function.
2. Halving the wavelength more than doubles the kinetic energy.

Answer: (b) 2 : 1

Explanation:
1. The threshold wavelength is the longest wavelength that works, so \( \phi = \dfrac{hc}{\lambda_0} \).
2. \( \dfrac{\phi_A}{\phi_B} = \dfrac{\lambda_B}{\lambda_A} = \dfrac{600}{300} = 2 \).

Teacher's Note:
1. Work function is inversely proportional to threshold wavelength.
2. The metal that responds to longer wavelengths has the smaller work function.

Answer: (b) the voltage applied to the tube

Explanation:
1. The most energetic photon is produced when an electron gives all its kinetic energy eV to one photon.
2. So \( \lambda_{min} = \dfrac{hc}{eV} = \dfrac{12400}{V}\ \text{Å} \), which depends only on the accelerating voltage.

Teacher's Note:
1. This is the Duane-Hunt law.
2. Current changes only the intensity, and the target changes only the characteristic lines.

Answer: (c) 2

Explanation:
1. Kinetic energy gained is \( K = qV \).
2. \( \dfrac{K_\alpha}{K_p} = \dfrac{2e\,V}{e\,V} = 2 \).

Teacher's Note:
1. Equal potential difference gives energy proportional to charge, not mass.
2. Equal kinetic energies would instead need different voltages.

Answer: (d) ratio of Planck's constant to the charge on the electron

Explanation:
1. Einstein's equation gives \( eV_0 = h\nu-\phi \), that is \( V_0 = \dfrac{h}{e}\nu-\dfrac{\phi}{e} \).
2. This is a straight line in V₀ against ν.
3. Its slope is \( \dfrac{h}{e} \).

Teacher's Note:
1. The intercept on the frequency axis is the threshold frequency.
2. Slope times e gives Planck's constant.