Dual Nature of Radiation and Matter — NEET UG practice

1. Recall that $E = h\nu$ gives the energy of a single light quantum, based on Planck's hypothesis — this represents the energy of a photon.

$$E = h\nu \;\rightarrow\; \text{IV. Energy of photon}$$

2. Since diffraction and interference occur only due to the superposition of wave amplitudes, they demonstrate light's wave nature.

$$\text{Diffraction \& Interference} \;\rightarrow\; \text{III. Wave nature of light}$$

3. The de Broglie relation connects a particle's wavelength to its momentum, describing matter's wave-like behaviour.

$$\lambda = \frac{h}{p} \;\rightarrow\; \text{I. de Broglie wavelength}$$

4. Since the Compton effect involves photons colliding with electrons and exchanging momentum like particles, it confirms light's particle nature.

$$\text{Compton effect} \;\rightarrow\; \text{II. Particle nature of light}$$

Hence, the correct matching is A-IV, B-III, C-I, D-II — option D.

1. Photoelectric effect fails to occur if the incident photon's energy is less than the work function, i.e., when the wavelength exceeds a certain threshold value.

$$\frac{hc}{\lambda} < W_0 \;\Rightarrow\; \lambda > \frac{hc}{W_0}$$

2. Substitute the given values of $h$, $c$, and $W_0 = 6.6\text{ eV}$ to find this threshold wavelength.

$$\lambda_{th} = \frac{6.6\times10^{-34}\times3\times10^8}{6.6\times1.6\times10^{-19}}$$

3. Simplify the expression.

$$\lambda_{th} = \frac{3\times10^{-7}}{1.6}\text{ m} = \frac{300}{1.6}\text{ nm} \approx 187.5\text{ nm}$$

4. Since the photoelectric effect fails only when $\lambda > 187.5\text{ nm}$, check which option exceeds this value. Among the given options, only 200 nm is greater than 187.5 nm.

Hence, the answer is option C.

1. Since both particles have the same energy, use the formula for a photon's momentum, which relates energy directly to the speed of light.

$$p_{Ph} = \frac{E_{Ph}}{c}$$

2. For the electron, since it is non-relativistic, use the kinetic energy–momentum relation.

$$p_e = \sqrt{2m_eE_e}$$

3. Given $E_e = E_{Ph} = 20\text{ eV}$, divide the two momenta to get the ratio.

$$\frac{p_e}{p_{Ph}} = \frac{\sqrt{2m_eE_e}}{E_{Ph}}\,c = c\sqrt{\frac{2m_e}{E_{Ph}}}$$

4. Substitute the given values, converting the energy to joules.

$$\frac{p_e}{p_{Ph}} = 3\times10^8 \times \sqrt{\frac{2\times9\times10^{-31}}{20\times1.6\times10^{-19}}}$$

5. Simplify the expression step by step.

$$\frac{p_e}{p_{Ph}} = \frac{3}{4}\times10^{-6}\times3\times10^8 = \frac{900}{4} = 225$$

Hence, the answer is option D.