Questions: Dual Nature Of Radiation And Matter

Q1. Define the work function of a metal. Answer (English): The work function is the minimum energy needed to remove an electron from the surface of a metal. A photon must have at least this energy for photoemission to occur, so the threshold frequency satisfies h nu0 equals the work function. Different metals have different work functions. Hinglish: Work function woh minimum energy hai jo metal ki surface se electron nikalne ke liye chahiye. h nu0 = phi se threshold frequency milti hai, aur har metal ka phi alag hota hai.

Q2. State Einstein's photoelectric equation. Answer (English): The maximum kinetic energy of the emitted electron equals the photon energy minus the work function, written as K maximum equals h nu minus phi. Equivalently, the stopping potential times the electron charge equals h nu minus phi. No emission occurs when the photon energy is less than the work function. Hinglish: Einstein equation: K_max = h nu - phi. Photon ki energy agar work function se kam ho to electron nikalta hi nahi.

Q3. How does the stopping potential depend on the intensity of incident light? Answer (English): The stopping potential does not depend on intensity. It depends only on the frequency of the light and on the metal, because it measures the maximum kinetic energy of the electrons. Increasing the intensity increases the number of emitted electrons and so the saturation current, but the fastest electrons still have the same kinetic energy. Hinglish: Stopping potential intensity par depend nahi karta. Sirf frequency aur metal par depend karta hai, kyunki yeh maximum kinetic energy naapta hai.

Q4. Why is a minimum frequency needed for photoelectric emission? Answer (English): A single photon must supply at least the work function of the metal to free an electron. Below the threshold frequency each photon carries less than the work function, so no single photon can eject an electron, however many photons arrive. A very bright beam below the threshold therefore produces no photocurrent. Hinglish: Har photon ko work function jitni energy chahiye. Threshold se neeche photon ki energy kam hoti hai, isliye intensity bada karne par bhi emission nahi hoti.

Q5. Write the de Broglie wavelength of a particle. Answer (English): The de Broglie wavelength is lambda equals h divided by p, where h is Planck's constant and p is the momentum of the particle. For a particle of mass m moving with speed v this becomes h divided by mv. A faster or heavier particle has a shorter wavelength. Hinglish: de Broglie wavelength lambda = h/p = h/(mv). Zyada speed ya zyada mass se wavelength chhoti hoti hai.

Q6. Write the de Broglie wavelength of an electron accelerated through a potential difference V. Answer (English): The kinetic energy of the electron is eV, so its momentum is the square root of twice m e V. The wavelength is therefore h divided by the square root of twice m e V. Numerically this is about 1.227 divided by the square root of V nanometres, with V in volts. Hinglish: Kinetic energy eV se momentum sqrt(2meV) milta hai, aur lambda = h/sqrt(2meV). Numerically lambda = 1.227/sqrt(V) nanometre.

Q7. What is the shortest wavelength limit of the continuous X-ray spectrum? Answer (English): The shortest wavelength in the continuous part of the X-ray spectrum corresponds to a photon that carries the entire kinetic energy of the incident electron. Its energy is therefore eV, and the shortest wavelength is hc divided by eV. This is called the cut-off wavelength of the spectrum. Hinglish: Jab poora electron energy ek photon me convert ho, photon ki energy eV hoti hai, isliye shortest wavelength hc/eV hoti hai. Ise cut-off wavelength kehte hain.

Q8. State the result of the Davisson and Germer experiment. Answer (English): They scattered a beam of electrons from a nickel crystal and observed strong maxima at particular angles. The angles matched the diffraction pattern expected from the de Broglie wavelength of the electrons. The experiment confirmed that particles have a wave nature. Hinglish: Davisson-Germer ne nickel crystal se electrons scatter kiye aur diffracted pattern dekha, jo de Broglie wavelength se match karta tha. Isse matter waves confirm hue.

Q9. How does the maximum kinetic energy of photoelectrons vary with the frequency of light? Answer (English): The maximum kinetic energy increases linearly with the frequency of the light. The graph of maximum kinetic energy against frequency is a straight line with slope h, the Planck constant, and the line meets the frequency axis at the threshold frequency. If the frequency doubles but stays above the threshold, the kinetic energy changes by the difference of the two photon energies, not by a simple doubling. Hinglish: K_max frequency ke saath linear badhti hai. Graph ka slope h hota hai aur frequency axis par threshold frequency par milta hai.

Q10. What is the photoelectric effect and why does it show that light behaves as a particle? Answer (English): The photoelectric effect is the emission of electrons from a metal surface when light of suitable frequency falls on it. Emission is instantaneous and depends on frequency rather than intensity, and a threshold frequency exists. These features fit the photon picture, in which one photon gives all its energy to one electron and no accumulation time is needed. Hinglish: Light metal surface par gir kar electrons nikaalta hai, ise photoelectric effect kehte hain. Instant emission aur threshold frequency photon picture ko support karte hain, kyunki ek photon apni poori energy ek electron ko deta hai.

Marks pattern for this chapter

One-mark items test the photon energy formula, the meaning of work function or the de Broglie relation. Two and three mark items ask for a stopping potential, a threshold wavelength, a wavelength ratio or the reading of a graph. Five-mark items combine the photoelectric equation with a graph or with the experimental observations. Writing the equation, the substitution in electron volts and the final unit covers each type.

Common mistakes to avoid

Using the wavelength in nanometres directly with h c in joule seconds gives wrong energy values; convert to electron volts or use h c in electron volt nanometres. Forgetting that the stopping potential is independent of intensity is a common conceptual slip. In de Broglie comparisons, confusing same speed with same energy changes the ratio by a square root. Finally, the work function belongs to the metal, never to the light.

In every de Broglie question, write the relation in terms of momentum first, then substitute whichever expression for momentum the given fixed quantity suggests.

Quiz below: five multiple choice questions on photon energy, the photoelectric equation, stopping potential and the de Broglie relation. Attempt them after revising the formula sheet and the two topic pages.