Question 1

Q1: 11.1 Find the (a) maximum frequency, and (b) minimum wavelength of X-rays produced by 30 kV electrons.

Accepted Answer

When an electron is accelerated by a potential difference V, the kinetic energy it gains is given by K.E. = eV. If this electron is stopped by a target, its kinetic energy can be converted into an X-ray photon. The maximum energy of the produced photon occurs when the entire kinetic energy of the el...

Question 2

Q10: 11.10 What is the de Broglie wavelength of (a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s, (b) a ball of mass 0.060 kg moving at a...

Accepted Answer

The de Broglie wavelength (λ) associated with a moving particle is given by the relation: λ = h/p = h/(mv) where h is Planck's constant, p is the momentum, m is the mass of the particle, and v is its speed. Given: - Planck's constant, h = 6.63 × 10⁻³⁴ J s (a) A bullet of mass 0.040 kg travelling at...

Question 3

Q11: 11.11 Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

Accepted Answer

This problem requires us to show that two different ways of defining wavelength—one from the classical wave theory of light and the other from the de Broglie hypothesis for particles—yield the same result for a photon. 1. Wavelength from the particle nature of light (de Broglie wavelength) Louis de...

Question 4

Q2: 11.2 The work function of caesium metal is 2.14 eV. When light of frequency 6 × 10¹⁴ Hz is incident on the metal surface, photoemission of electrons o...

Accepted Answer

Given: - Work function of caesium, φ₀ = 2.14 eV - Frequency of incident light, ν = 6 × 10¹⁴ Hz - Planck's constant, h = 6.63 × 10⁻³⁴ J s - Charge of an electron, e = 1.602 × 10⁻¹⁹ C - Mass of an electron, m = 9.11 × 10⁻³¹ kg - 1 eV = 1.602 × 10⁻¹⁹ J First, let us calculate the energy of the incident...

Question 5

Q3: 11.3 The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?

Accepted Answer

The photoelectric cut-off voltage, also known as the stopping potential (V₀), is the minimum negative potential applied to the collector plate that stops the photoelectric current completely. This potential is just sufficient to repel the most energetic photoelectrons. The relationship between the s...

Question 6

Q4: 11.4 Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW. (a) Find the energy and momentum of...

Accepted Answer

Given: - Wavelength, λ = 632.8 nm = 632.8 × 10⁻⁹ m - Power emitted, P = 9.42 mW = 9.42 × 10⁻³ W - Planck's constant, h = 6.63 × 10⁻³⁴ J s - Speed of light, c = 3 × 10⁸ m/s - Mass of a hydrogen atom, m_H ≈ 1.67 × 10⁻²⁷ kg (a) Energy and momentum of each photon    Energy (E):     The energy of a singl...

Question 7

Q5: 11.5 In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 × 10⁻¹⁵ V s. Ca...

Accepted Answer

According to Einstein's photoelectric equation, the maximum kinetic energy (K_max) of an emitted photoelectron is given by: K_max = hν - φ₀ where h is Planck's constant, ν is the frequency of the incident light, and φ₀ is the work function of the metal. The maximum kinetic energy is also related to...

Question 8

Q6: 11.6 The threshold frequency for a certain metal is 3.3 × 10¹⁴ Hz. If light of frequency 8.2 × 10¹⁴ Hz is incident on the metal, predict the cutoff vo...

Accepted Answer

The cut-off voltage (or stopping potential, V₀) for photoelectric emission can be predicted using Einstein's photoelectric equation. The equation is: K_max = hν - φ₀ where K_max is the maximum kinetic energy of the photoelectrons, h is Planck's constant, ν is the frequency of incident light, and φ₀...

Question 9

Q7: 11.7 The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?

Accepted Answer

For photoelectric emission to occur, the energy of the incident photons (E) must be greater than or equal to the work function (φ₀) of the metal. If the photon energy is less than the work function, no electrons will be emitted. Condition for photoelectric emission: E ≥ φ₀ Given: - Work function, φ₀...

Question 10

Q8: 11.8 Light of frequency 7.21 × 10¹⁴ Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 10⁵ m/s are ejected from the surface. W...

Accepted Answer

To find the threshold frequency (ν₀), we first need to determine the work function (φ₀) of the metal using Einstein's photoelectric equation: K_max = hν - φ₀ where K_max is the maximum kinetic energy of the emitted electrons, h is Planck's constant, and ν is the frequency of the incident light. Give...

Question 11

Q9: 11.9 Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is inciden...

Accepted Answer

The work function (φ₀) of a material can be determined using Einstein's photoelectric equation, which relates the energy of the incident photon, the work function, and the maximum kinetic energy of the emitted electrons. The equation is: E = φ₀ + K_max where E is the energy of the incident photon an...

NCERT Solutions