NEET Dual Nature of Radiation and Matter, Atoms and Nuclei: Important Questions by PW

The three chapters Dual Nature of Radiation and Matter, Atoms, and Nuclei connect some of the most important ideas in modern physics. Questions often require you to identify the correct formula, interpret a graph, calculate an energy difference or predict how a nucleus changes during decay.

Use Physics Wallah’s Dual Nature of Radiation and Matter, Atoms and Nuclei Important Questions to practise numerical problems, revise frequently tested relationships and strengthen your understanding of concepts that are easy to confuse.

Important Questions

Q1. What is the de Broglie wavelength of a moving particle?

Answer: The de Broglie wavelength is the wavelength associated with a moving material particle.

The formula is:

λ = h/p

For a non-relativistic particle, momentum is p = mv. Therefore:

λ = h/mv

Where:

• λ = de Broglie wavelength

• h = Planck’s constant = 6.626 × 10⁻³⁴ J s

• p = momentum

• m = mass of the particle

• v = velocity of the particle

If the kinetic energy K is given, use:

λ = h/√(2mK)

Key point: A particle’s wavelength decreases as its momentum increases.

Q2. How do you calculate the de Broglie wavelength of a charged particle accelerated through a potential difference?

Answer: A charged particle accelerated through a potential difference V gains kinetic energy.

For a particle of mass m and charge magnitude q, assuming it starts from rest:

qV = K = ½mv²

Using λ = h/√(2mK):

λ = h/√(2mqV)

This formula applies to a non-relativistic particle accelerated from rest through a potential difference V.

Q3. What is the de Broglie wavelength of an electron accelerated through a potential difference V?

Answer: For an electron accelerated through V volts, the approximate formula is:

λ = 12.27/√V Å

Here, V is the accelerating potential in volts and Å represents angstroms.

Example: An electron is accelerated through 100 V. Find its de Broglie wavelength.

λ = 12.27/√100 Å

λ = 12.27/10 Å

λ = 1.227 Å

This is a useful relation for numerical questions involving electrons.

Q4. What is the de Broglie wavelength of a gas molecule at temperature T?

Answer: For a gas molecule whose kinetic energy is taken as the average translational kinetic energy,

K = 3kT/2

Substituting into λ = h/√(2mK):

λ = h/√(3mkT)

Where k is Boltzmann’s constant and T is the absolute temperature in kelvin.

Remember: Select the formula according to the kinetic energy specified in the question. The expression above uses average translational kinetic energy, not the kinetic energy of every individual molecule.

Q5. What is the difference between the wavelength relations for matter particles and photons?

Answer: Both matter particles and photons have a de Broglie wavelength given by:

λ = h/p

For a non-relativistic material particle:

λ = h/mv

For a photon:

E = pc = hc/λ

Therefore:

λ = hc/E

A photon has zero rest mass but carries energy and momentum.

Q6. What is radiation pressure? How does it differ for absorbing and reflecting surfaces?

Answer: Radiation pressure is the force exerted by electromagnetic radiation per unit area on a surface.

For radiation of intensity I incident normally on an area A:

Here, c is the speed of light.

Key point: For normal incidence, a perfectly reflecting surface experiences twice the radiation pressure of a perfectly absorbing surface under the same incident intensity.

Q7. How do you calculate the number of photons incident per unit time?

Answer: If radiation of intensity I falls normally on an area A, the incident power is:

P = IA

Energy of one photon:

E = hν = hc/λ

If N represents the number of photons incident per unit time, then:

N = P/hν

Therefore:

N = IA/hν = IAλ/hc

You can also use:

N = Pλ/hc

Example: A light source delivers power P at frequency ν. The number of photons arriving per second is P divided by hν.

Remember that N here means a photon rate, measured in photons per second.

Q8. What is the photoelectric effect?

Answer: The photoelectric effect is the emission of electrons from a material surface when incident electromagnetic radiation has a sufficiently high frequency.

The emitted electrons are called photoelectrons, and the resulting flow of charge is measured as photoelectric current.

The effect demonstrates the particle nature of light because the energy transferred to an electron depends on the energy of an individual photon.

Q9. What is threshold frequency? When does photoelectric emission occur?

Answer: Threshold frequency, ν₀, is the minimum frequency required to cause photoelectric emission from a particular material.

• If ν > ν₀, photoelectric emission occurs.

• If ν = ν₀, emission can begin, with zero maximum kinetic energy in the ideal case.

• If ν < ν₀, photoelectric emission does not occur, regardless of the intensity of the incident light.

The work function is:

φ = hν₀

The threshold frequency depends on the material.

Q10. State Einstein’s photoelectric equation.

Answer: Einstein’s photoelectric equation follows conservation of energy. The energy of an incident photon is used to overcome the work function, and the remaining energy becomes the maximum kinetic energy of the emitted electron.

hν = φ + Kₘₐₓ

Since φ = hν₀:

Kₘₐₓ = hν − hν₀

The work function φ is the minimum energy required to remove an electron from the material surface.

Memory tip: Photon energy = Work function + Maximum kinetic energy.

Q11. What are the important observations of the photoelectric effect?

Answer:

• Photoelectric emission occurs only when the incident frequency is at or above the threshold frequency.

• Increasing the frequency above threshold increases the maximum kinetic energy of the emitted electrons.

• Increasing intensity at a fixed frequency above threshold generally increases the photoelectric current.

• Photoelectric emission occurs with negligible time delay under ordinary experimental conditions.

• Emitted electrons can have different kinetic energies because they originate from different energy states and may lose energy inside the material.

Q12. What is stopping potential? How is it related to maximum kinetic energy?

Answer: Stopping potential, Vₛ, is the minimum reverse potential needed to stop even the fastest photoelectrons from reaching the collector.

The relationship is:

Kₘₐₓ = eVₛ

Using Einstein’s equation:

eVₛ = hν − φ

Therefore:

Vₛ = (h/e)ν − φ/e

Where e is the magnitude of the electronic charge.

Q13. What does the graph of stopping potential against frequency show?

Answer: The graph of stopping potential Vₛ against incident frequency ν is a straight line.

From:

Vₛ = (h/e)ν − φ/e

• Slope: h/e

• Frequency-axis intercept: ν₀ = φ/h

• Stopping-potential-axis intercept: −φ/e

For different metals, the slopes are the same in the ideal Einstein model because h/e is a universal constant. Metals with larger work functions have higher threshold frequencies and more negative stopping-potential-axis intercepts.

Q14. What is cut-off wavelength? How is it related to work function?

Answer: Cut-off wavelength, λ₀, is the maximum wavelength that can produce photoelectric emission from a given material.

Since frequency and wavelength are related by c = νλ:

φ = hν₀ = hc/λ₀

The photon energy is:

E = hν = hc/λ

A wavelength longer than λ₀ corresponds to photon energy below the work function, so photoelectric emission does not occur.

Q15. How can you calculate the work function or stopping potential from two photoelectric conditions?

Answer: Write Einstein’s equation for each condition.

For two incident frequencies ν₁ and ν₂:

eVₛ₁ = hν₁ − φ

eVₛ₂ = hν₂ − φ

Subtracting eliminates the common work function:

e(Vₛ₂ − Vₛ₁) = h(ν₂ − ν₁)

Therefore:

h/e = (Vₛ₂ − Vₛ₁)/(ν₂ − ν₁)

This method is useful when two stopping potentials or two photoelectric conditions are given.

Atoms: Important Questions

Q16. What was Thomson’s atomic model, and why was it rejected?

Answer: Thomson proposed that an atom consisted of a positively charged region with electrons embedded in it.

The model could not explain the results of Rutherford’s alpha-particle scattering experiment, particularly the large deflections experienced by a very small fraction of alpha particles. It was therefore replaced by the nuclear model of the atom.

Q17. What were the observations and conclusions of Rutherford’s gold-foil experiment?

Answer: Rutherford directed alpha particles at a thin gold foil and observed their scattering.

Memory tip: Straight passage indicates empty space; large-angle scattering indicates a small, dense, positively charged nucleus.

Q18. Why did Rutherford’s atomic model fail to explain atomic stability?

Answer: According to classical electromagnetic theory, a revolving electron is an accelerating charged particle and should continuously radiate energy.

As it lost energy, the electron would move towards the nucleus. Rutherford’s model could not explain why atoms remain stable or why they produce discrete line spectra.

Q19. State Bohr’s postulates for the hydrogen atom.

Answer: Bohr proposed that electrons can occupy certain stable orbits without continuously radiating energy.

The main relations are:

Quantisation of angular momentum:

mvr = nh/2π

where n = 1, 2, 3, …

Energy of the nth orbit in a hydrogen-like atom:

Eₙ = −13.6Z²/n² eV

Radius of the nth orbit:

rₙ = 0.529n²/Z Å

Electron speed in the nth orbit:

vₙ ≈ 2.18 × 10⁶(Z/n) m/s

Here, Z is the atomic number of the hydrogen-like species.

An electron emits or absorbs a photon when it transitions between two permitted energy levels.

Q20. How is photon energy related to an electron’s transition between two Bohr orbits?

Answer: The energy of the photon equals the magnitude of the energy difference between the two levels.

For emission from a higher energy level nᵢ to a lower level n𝒻:

hν = Eᵢ − E𝒻

The photon wavelength is given by:

1/λ = RZ²(1/n𝒻² − 1/nᵢ²)

where nᵢ > n𝒻 and R is the Rydberg constant.

For absorption, the electron moves from a lower level to a higher level and absorbs a photon whose energy equals the energy difference.

Q21. How many spectral lines can be produced when an electron transitions from an excited level n to lower levels?

Answer: If all possible downward transitions among the first n energy levels are considered, the maximum number of distinct spectral lines is:

N = n(n − 1)/2

Example: If an electron is initially in the fourth energy level:

N = 4(4 − 1)/2

N = 4 × 3/2

N = 6 spectral lines

This formula assumes that all possible transitions between the relevant levels can occur.

Q22. How are kinetic energy, potential energy and total energy related in a Bohr orbit?

Answer: For an electron in a Coulomb-bound Bohr orbit:

• Potential energy is negative.

• Kinetic energy is positive.

• Total energy is negative.

The relationships are:

U = −2K

E = K + U = −K = U/2

Example: If the potential energy is −8 eV:

K = −U/2 = +4 eV

E = K + U = 4 − 8

E = −4 eV

Q23. What is the ionisation energy of a hydrogen-like atom in its nth orbit?

Answer: The ionisation energy is the energy required to remove the electron completely from that orbit.

Since:

Eₙ = −13.6Z²/n² eV

The ionisation energy from the nth orbit is:

Eᵢₒₙ = 13.6Z²/n² eV

For the ground state, n = 1:

Eᵢₒₙ = 13.6Z² eV

For hydrogen, Z = 1, so the ground-state ionisation energy is 13.6 eV.

Nuclei: Important Questions

Q24. How do you determine the number of protons and neutrons in a nucleus?

Answer: A nucleus is represented as ᴬZX, where A is the mass number and Z is the atomic number.

• Number of protons = Z

• Number of neutrons = A − Z

• Mass number = number of protons + number of neutrons

Example: For ²³₁₁Na:

Protons = 11

Neutrons = 23 − 11 = 12

Q25. How does nuclear radius depend on mass number?

Answer: Nuclear radius is approximately related to mass number by:

R = R₀A¹ᐟ³

where R₀ is approximately 1.2 fm and A is the mass number.

Since nuclear volume is proportional to R³, it is proportional to A. Nuclear mass is also approximately proportional to A. Therefore, nuclear density is approximately the same for different nuclei.

Q26. What is mass defect? How is binding energy calculated?

Answer: Mass defect is the difference between the sum of the masses of the separate constituent nucleons and the actual mass of the bound nucleus.

Δm = (sum of free nucleon masses) − (actual nuclear mass)

Binding energy is the energy required to separate the nucleus completely into its constituent nucleons.

BE = Δmc²

If the mass defect is expressed in atomic mass units:

BE (MeV) = Δm (u) × 931.5

Q27. How is nuclear stability compared using binding energy per nucleon?

Answer: Binding energy per nucleon is the average binding energy associated with each nucleon.

Binding energy per nucleon = Total binding energy/A

A larger binding energy per nucleon generally indicates a more tightly bound nucleus and greater stability against separation into individual nucleons.

Nuclei around the iron-nickel region have among the highest binding energies per nucleon. This helps explain why energy can be released by fission of very heavy nuclei and fusion of sufficiently light nuclei.

Q28. How do alpha, beta and gamma decay affect atomic number and mass number?

Answer: The changes depend on the type of radioactive decay.

Examples:

Alpha decay:

ᴬZX → ᴬ⁻⁴Z⁻²Y + ⁴₂He

Beta-minus decay:

ᴬZX → ᴬZ+1Y + e⁻ + antineutrino

Beta-plus decay:

ᴬZX → ᴬZ−1Y + e⁺ + neutrino

Gamma emission releases energy from an excited nucleus without changing its atomic number or mass number.

Q29. What happens to a nucleus during alpha decay?

Answer: An alpha particle contains two protons and two neutrons. When a nucleus emits an alpha particle:

• Its atomic number decreases by 2.

• Its mass number decreases by 4.

The released energy is shared between the alpha particle and the recoiling daughter nucleus, with the exact distribution depending on the decay and conservation of momentum.

Q30. What happens during beta-minus decay, beta-plus decay and electron capture?

Answer:

• Beta-minus decay: A neutron changes into a proton, emitting an electron and an antineutrino. The atomic number increases by 1.

• Beta-plus decay: A proton changes into a neutron, emitting a positron and a neutrino. The atomic number decreases by 1.

• Electron capture: The nucleus captures an inner-shell electron. A proton changes into a neutron, and the atomic number decreases by 1.

In all three processes, the mass number remains unchanged.

Revise these questions by first identifying the concept being tested, then selecting the relevant formula and checking the units before calculating. Pay particular attention to photoelectric graphs, Bohr energy transitions, radioactive decay and binding energy. Physics Wallah’s Physics Important Questions can support your preparation by helping you practise these concepts and strengthen your numerical problem-solving.

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