Previous year questions: Dual Nature Of Radiation And Matter
Dual Nature of Radiation and Matter28 min readLesson 5 of 5
Q1 (CBSE Board 2022, 2 marks)
A metal surface is irradiated with frequencies v1 and v2, and the maximum kinetic energies of the emitted electrons are K and 2K respectively. Obtain the expression for the threshold frequency of the surface. Answer: Write K equal to h v1 minus the work function, and 2K equal to h v2 minus the work function. Subtracting the first from the second gives K equal to h times (v2 minus v1). Substituting this K back in the first equation gives the work function equal to h times v1 minus h times (v2 minus v1), which is h times (2 v1 minus v2). A quick check: if v2 equals three times v1, the threshold frequency becomes v1, which lies below both incident frequencies. The threshold frequency is the work function divided by h, so it is 2 v1 minus v2.
Q2 (CBSE Board 2022, 2 marks)
An alpha particle and a proton have the same speed and then the same kinetic energy. Find the ratio of their de Broglie wavelengths in each case. Answer: For the same speed the de Broglie wavelength is inversely proportional to the mass, so the ratio of the alpha particle wavelength to the proton wavelength is the inverse ratio of their masses, which is 1 to 4. For the same kinetic energy the wavelength is inversely proportional to the square root of the mass, so the ratio is the inverse of the square root of 4, which is 1 to 2. An alpha particle has twice the charge and four times the mass of a proton, which is where the factors of two and four come from. Writing the mass ratio as four to one before taking the square root avoids the common inversion error.
Q3 (CBSE Board 2022, 3 marks)
Use Einstein's photoelectric equation to draw the variation of the maximum kinetic energy of the emitted electrons against the frequency of the incident light, and explain the intercept on each axis. Answer: The graph is a straight line. The slope equals the Planck constant, because each extra unit of frequency adds h of energy. The intercept on the energy axis is minus the work function, and the intercept on the frequency axis is the threshold frequency. For frequencies below that point the maximum kinetic energy would be negative, which means no emission. The same graph can also be used to read the work function, because the intercept on the energy axis is minus phi. The graph is linear for every metal, with only the intercept changing from one metal to another.
Q4 (CBSE Board 2022, 2 marks)
What are matter waves? A proton and an alpha particle are accelerated through the same potential difference. Find the ratio of their de Broglie wavelengths. Answer: Matter waves are the waves associated with moving particles, with wavelength h divided by momentum. Accelerated through the same potential, a particle gains kinetic energy qV, so its momentum is the square root of twice m q V and its wavelength is h divided by that square root. The ratio of the proton wavelength to the alpha particle wavelength is therefore the square root of (m alpha q alpha) divided by (m proton q proton), which is the square root of 4 times 2 divided by 1 times 1, equal to the square root of 8. With a mass ratio of four and a charge ratio of two, the proton wavelength comes out two root two times the alpha particle wavelength. So the proton wavelength is 2 root 2 times the alpha particle wavelength.
Q5 (CBSE Board 2023, 1 mark)
A photon of energy 7 eV is incident on a metal surface having a work function of 3.75 eV. Which stopping potential is required: 1.25 V, 2.45 V, 3.25 V or 3.75 V? Answer: The maximum kinetic energy is the photon energy minus the work function, so 7 minus 3.75 equals 3.25 eV. The stopping potential in volts has the same numerical value as the maximum kinetic energy in electron volts, so it is 3.25 V. The correct choice is 3.25 V. The saturation current in the same experiment would be set by the intensity, not by the energy of each photon. The photon energy and the work function are both in electron volts here, so the subtraction is direct.
Q6 (CBSE Board 2023, 1 mark)
In an experiment, photons of energy 7.5 eV are incident on a metal surface, and the emitted electrons are stopped by a potential of 4.5 V. What is the work function: 3.0 eV, 4.5 eV, 7.5 eV or 12.0 eV? Answer: The maximum kinetic energy in electron volts equals the stopping potential in volts, so it is 4.5 eV. The work function is the photon energy minus this kinetic energy, which is 7.5 minus 4.5, equal to 3.0 eV. The correct choice is 3.0 eV. The threshold wavelength of this metal follows from the work function equal to 1240 divided by the threshold wavelength, which is about 413 nm. The work function is a property of the surface, so it does not change with the light used.
Q7 (CBSE Board 2023, 2 marks)
A particle is associated with a de Broglie wave of wavelength 1.2 nm. If its kinetic energy is made four times larger, what is the new de Broglie wavelength? Answer: The de Broglie wavelength is inversely proportional to the square root of the kinetic energy. Making the kinetic energy four times larger multiplies the wavelength by one over the square root of four, which is one half. The new wavelength is 1.2 divided by 2, equal to 0.6 nm. Four times the kinetic energy doubles the momentum, so the wavelength becomes half of 1.2 nm, that is 0.6 nm. The square root is the only place where the factor of four enters.
Q8 (CBSE Board 2023, 1 mark)
Assertion: the photoelectric effect demonstrates the particle nature of light. Reason: the photoelectric current is proportional to the intensity of the incident radiation for frequencies above the threshold. Choose the correct relation between assertion and reason. Answer: Both statements are true. The assertion is the consequence of the photon picture, in which one photon interacts with one electron. The reason describes the observed dependence of the saturation current on intensity, which is also explained by the photon picture. However the reason is not the explanation of the assertion, so the correct choice is that both are true but the reason is not the correct explanation. The second statement is about the number of electrons and does not explain the particle claim. The assertion follows because a single photon transfers its whole energy to one electron in one event.
Q9 (CBSE Board 2024, 1 mark)
Two beams with photon energies 3.3 eV and 11.3 eV illuminate a metal surface of work function 2.3 eV one after the other. Find the ratio of the maximum speeds of the emitted electrons. Answer: The maximum kinetic energies are 3.3 minus 2.3 equals 1.0 eV and 11.3 minus 2.3 equals 9.0 eV. The speed is proportional to the square root of the kinetic energy, so the ratio of the speeds is the square root of 1 to 9, which is 1 to 3. The slower beam gives one third of the speed. The maximum kinetic energies are 1.0 eV and 9.0 eV respectively, and the speed ratio is the square root of the ratio of these energies. The squares of the speeds are in the ratio of the kinetic energies.
Q10 (CBSE Board 2024, 2 marks)
Consider a neutron of kinetic energy E and a photon of the same energy. Obtain the relation between the de Broglie wavelength of the neutron and the wavelength of the photon. Answer: For the neutron the de Broglie wavelength is h divided by the square root of twice m E, where m is the neutron mass. For the photon the energy E equals h c divided by its wavelength, so its wavelength is h c divided by E. Dividing the two gives the neutron wavelength equal to the photon wavelength times the square root of (E divided by twice m c squared), which shows the photon wavelength is much larger for the same energy. At the same energy the photon momentum is E divided by c while the neutron momentum is the square root of twice m E, so the photon wavelength is the larger one. Both wavelengths contain the energy, so the ratio reduces to the square root of the mass term.