NEET Dual Nature of Radiation and Matter Previous Year Questions with Solutions

The number of incident photons per second is given by n = P(t) / (h*nu). Since each absorbed photon can eject at most one photoelectron (with quantum efficiency eta):

• • Because power P(t) decreases linearly with time (P(t) = P_0 - k*t), photocurrent I must also decrease linearly with time.
• • The graph of I vs t is a straight line sloping downward to zero.

Apply Einstein's photoelectric equation for maximum kinetic energy of emitted electrons:

• • Stopping potential depends solely on the frequency nu of incident radiation and the work function Phi.
• • Stopping potential is completely INDEPENDENT of light intensity and source power.
• • Therefore, |V| vs t is a horizontal straight line with zero slope.

Inspect the four graph sets in the question crop:

• • Matches Set (1) perfectly, corresponding to Option (1).

For photoelectric emission to occur, incident photon energy must equal or exceed the work function:

• • For Cell 1, we are given lambda_1 < lambda, which means lambda > lambda_1.
• • The incident wavelength exceeds threshold wavelength; photon energy is strictly below the work function.
• • No photoelectrons are ejected, so the stopping potential is identically zero: V_1 = 0.

For Cells 2 and 3, lambda < lambda_2 and lambda << lambda_3, so photoelectric emission takes place in both cells:

• • eV_2 = hc (1/lambda - 1/lambda_2).
• • eV_3 = hc (1/lambda - 1/lambda_3).
• • Given lambda_3 >> lambda_2 > lambda, we have 1/lambda_3 < 1/lambda_2.
• • Subtracting a smaller number yields a larger difference: (1/lambda - 1/lambda_3) > (1/lambda - 1/lambda_2).

Compare the stopping potential magnitudes:

• • Combining with V_1 = 0 yields: V_1 = 0, V_2 < V_3.
• • Matches Option (1) exactly.

For a massless photon of energy E traveling at speed c:

• • Given E = 20 eV = 20 * 1.6 * 10^-19 J = 3.2 * 10^-18 J.
• • p_{Ph} = (3.2 * 10^-18 J) / (3 * 10^8 m/s) = (3.2 / 3) * 10^-26 kg m/s.

For an electron with kinetic energy E = 20 eV, non-relativistic momentum is:

• • Inside square root: 2 * 9 * 3.2 * 10^-49 = 57.6 * 10^-49 = 5.76 * 10^-48.
• • Taking square root: sqrt(5.76 * 10^-48) = 2.4 * 10^-24 kg m/s.

Divide electron momentum by photon momentum:

• • Confirms the exact integer ratio 225.
• • Matches Option (3) exactly.

Louis de Broglie explained Bohr’s angular momentum quantization by proposing that an electron in an orbit behaves as a standing matter wave. Constructive interference requires the orbital circumference to contain an integral number of de Broglie wavelengths:

• • For n = 2: λ = 2π r₂ / 2 = π r₂.

The radius of the n-th Bohr orbit scales with the square of the principal quantum number:

• • r₂ = 0.208 nm (or 0.212 nm with standard a₀ = 0.0529 nm).

Substitute r₂ into the wavelength relation:

• • Matches Option (2) 0.67 nm.

For electromagnetic radiation (photons), energy is related to wavelength by the Planck-Einstein relation:

• • c is the speed of light in vacuum.

For a massive non-relativistic particle of mass m and kinetic energy E, momentum is p = √(2 m E):

• • This is de Broglie’s fundamental relation for matter waves.

Divide the photon wavelength by the electron wavelength:

• • Matches Option (3) exactly.

Intensity of light represents the energy incident per unit area per second, which for a given frequency is directly proportional to the number of incident photons per second. Since one photon liberates at most one photoelectron:

• • Thus, photoelectric current increases linearly with light intensity, starting from zero at zero intensity.
• • Graph A accurately represents this linear proportionality.

For incident light with frequency ν greater than the threshold frequency ν₀, changing the frequency while maintaining constant intensity does NOT change the number of incident photons (in fact, if energy intensity is constant, photon flux I/(hν) slightly decreases). In standard NCERT experiments where photon flux is maintained:

• • Photoelectric current does NOT increase linearly with frequency.
• • Hence, Graphs C and D are completely incorrect for photoelectric current.

Graph D shows a linear relationship with a non-zero threshold intercept:

• • This straight line with intercept ν₀ represents Stopping Potential V₀ vs. Frequency ν, NOT Photoelectric Current!
• • Therefore, Graph A is the single and only correct representation, corresponding to Option (1).

According to Planck’s quantum hypothesis as formulated by Einstein, electromagnetic radiation is emitted and absorbed in discrete packets called quanta or photons. The energy of a photon of frequency ν is given by:

In free space (vacuum), all photons, irrespective of their frequency or wavelength, travel at the universal constant speed of light c = 3 × 10⁸ m/s:

From relativistic kinematics and the de Broglie relation, the momentum p of a photon is related to its energy E by p = E/c:

In particle-like collisions such as photon-electron collisions (e.g. Compton scattering), the interaction is perfectly elastic, meaning both total energy and total momentum are conserved:

Photons are electrically neutral. They are not deflected by electric or magnetic fields and possess zero rest mass and zero electrical charge:

For a non-relativistic free particle of mass m moving with momentum p and kinetic energy E, the momentum is expressed in terms of kinetic energy as p = √(2mE). The de Broglie wavelength is:

Squaring both sides of the relation:

Taking the reciprocal gives the function plotted on the vertical axis (1/λ²) in terms of the horizontal axis variable E:

Since 2m/h² is a positive constant, the relation has the mathematical form y = k · x with zero intercept (c = 0). The corresponding graph is a straight line passing through the origin with positive slope.

An electron of charge e accelerated from rest through potential difference V acquires kinetic energy:

In continuous X-ray production (Bremsstrahlung), the minimum wavelength λ_min corresponds to the maximum photon energy, occurring when the entire kinetic energy of the electron is transferred to a single photon in a single collision:

Solve for λ_min in terms of potential difference V:

According to Einstein's quantum theory of the photoelectric effect, an incident photon transfers its entire energy E = hν to a single conduction electron. Photoelectric emission can occur only if this energy exceeds the minimum energy required to liberate the electron from the metal surface (the work function Φ₀):

Compare the incident radiation energy E = 2.20 eV with the work functions of all three photosensitive alkali metals:

Only Caesium satisfies the condition E ≥ Φ₀, meaning photoelectrons are emitted from the Caesium surface only.

According to Planck's quantum theory of radiation, light consists of discrete energy quanta called photons. The energy $E$ carried by a photon of wavelength $\lambda = 600\text{ nm} = 6 \times 10^{-7}\text{ m}$ is:

If the monochromatic source emits $n$ photons per second on average, the total energy radiated per second equals the source power $P$:

Substitute $P = 3.3 \times 10^{-3}\text{ W}$ and $E = 3.3 \times 10^{-19}\text{ J}$:

For incident photons of wavelength $\lambda$ incident on a surface of negligible work function ($\phi_0 \approx 0$), the maximum kinetic energy $K$ of emitted photoelectrons is:

The de-Broglie wavelength $\lambda_d$ of a particle of mass $m$ and kinetic energy $K$ is given by:

Substitute $K = \frac{hc}{\lambda}$ into the expression for $\lambda_d^2$:

Multiply both sides by $\frac{2mc}{h}$:

According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is: $$K_{\text{max}} = h\nu - \phi_0 = h(\nu - \nu_0)$$ For any photoelectrons to be emitted, the energy of the incident photon must be at least equal to the work function ($$h\nu \ge h\nu_0 \implies \nu \ge \nu_0$$).

The initial frequency of incident light is: $$\nu = 1.5 \nu_0$$ When the frequency is halved: $$\nu' = \frac{\nu}{2} = \frac{1.5\nu_0}{2} = 0.75\nu_0$$

Notice that the new frequency $$\nu' = 0.75\nu_0$$ is strictly LESS than the threshold frequency $$\nu_0$$: $$\nu' < \nu_0$$ When the frequency is below the threshold frequency, the photon energy is insufficient to liberate even a single electron from the surface.

Intensity corresponds to the number of incident photons per second. Since each individual photon carries energy less than the work function ($$h\nu' < \phi_0$$), increasing the number of photons cannot cause emission. Hence, no photoelectrons are ejected, and the photoelectric current drops to exactly ZERO.

One electron-volt ($$1\text{ eV}$$) is defined as the amount of kinetic energy gained or lost by a single electron when accelerated through an electrostatic potential difference of one volt: $$1\text{ eV} = e \times 1\text{ V} = 1.6 \times 10^{-19}\text{ J}$$.

Given the bond-dissociation energy $$E = 10^{-20}\text{ J}$$, express this energy in $$\text{eV}$$: $$E(\text{eV}) = \frac{E(\text{Joules})}{1.6 \times 10^{-19}\text{ J/eV}}$$.

Separate powers of ten: $$\frac{10^{-20}}{10^{-19}} = 10^{-20 - (-19)} = 10^{-1} = 0.1$$. Now divide by the coefficient: $$E = \frac{0.1}{1.6} = \frac{1}{16} = 0.0625\text{ eV}$$.

The calculated energy is $$0.0625\text{ eV}$$, which is nearly $$0.06\text{ eV}$$, matching Option 3.

An electron of mass $$m$$ and charge $$e$$ accelerated from rest through potential difference $$V$$ acquires kinetic energy $$K = eV$$. Its de Broglie wavelength is: $$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m e V}}$$.

Substituting $$h = 6.626 \times 10^{-34}\text{ J s}$$, $$m = 9.1 \times 10^{-31}\text{ kg}$$, and $$e = 1.6 \times 10^{-19}\text{ C}$$ gives the standard compact formula: $$\lambda = \frac{1.227}{\sqrt{V}}\text{ nm} = \frac{12.27}{\sqrt{V}}\text{ \AA}$$.

Given $$\lambda = 1.227 \times 10^{-2}\text{ nm}$$: $$1.227 \times 10^{-2} = \frac{1.227}{\sqrt{V}}$$ Divide both sides by $$1.227$$: $$10^{-2} = \frac{1}{\sqrt{V}} \implies \sqrt{V} = 10^2 = 100$$. Squaring both sides: $$V = 100^2 = 10000\text{ V} = 10^4\text{ V}$$.

Matches Option 4.