Dual Nature Of Radiation And MatterClass 12 Physics NCERT Solutions

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 electron is converted into a single photon. This corresponds to the maximum frequency (ν_max) and minimum wavelength (λ_min) of the X-rays.

Given:

The maximum energy of the photon is equal to the kinetic energy of the electron: E_max = eV We also know that the energy of a photon is E = hν. Therefore, hν_max = eV ν_max = eV / h

ν_max = (1.602 × 10⁻¹⁹ C × 30 × 10³ V) / (6.63 × 10⁻³⁴ J s) ν_max = (4.806 × 10⁻¹⁵ J) / (6.63 × 10⁻³⁴ J s) ν_max ≈ 7.25 × 10¹⁸ Hz

The relationship between photon energy and wavelength is E = hc/λ. For maximum energy, the wavelength is minimum: E_max = hc / λ_min Since E_max = eV, λ_min = hc / eV

λ_min = (6.63 × 10⁻³⁴ J s × 3 × 10⁸ m/s) / (1.602 × 10⁻¹⁹ C × 30 × 10³ V) λ_min = (19.89 × 10⁻²⁶ J m) / (4.806 × 10⁻¹⁵ J) λ_min ≈ 4.14 × 10⁻¹¹ m = 0.0414 nm

Thus, the maximum frequency is 7.25 × 10¹⁸ Hz and the minimum wavelength is 0.0414 nm.

Given:

First, let us calculate the energy of the incident photon (E) in eV. E = hν E = (6.63 × 10⁻³⁴ J s) × (6 × 10¹⁴ Hz) = 3.978 × 10⁻¹⁹ J To convert this energy to eV, we divide by the charge of an electron: E (in eV) = (3.978 × 10⁻¹⁹ J) / (1.602 × 10⁻¹⁹ J/eV) ≈ 2.48 eV

According to Einstein's photoelectric equation: K_max = E - φ₀ K_max = 2.48 eV - 2.14 eV K_max = 0.34 eV

To express this in Joules: K_max = 0.34 eV × 1.602 × 10⁻¹⁹ J/eV ≈ 5.45 × 10⁻²⁰ J

The stopping potential is the potential required to stop the most energetic electrons. It is related to the maximum kinetic energy by: K_max = eV₀ Therefore, V₀ = K_max / e If K_max is expressed in eV, the stopping potential in Volts is numerically equal to it. So, V₀ = 0.34 V

The maximum kinetic energy is also given by: K_max = (1/2)mv_max² We use the value of K_max in Joules: 5.45 × 10⁻²⁰ J = (1/2) × (9.11 × 10⁻³¹ kg) × v_max² v_max² = (2 × 5.45 × 10⁻²⁰) / (9.11 × 10⁻³¹) v_max² ≈ 1.196 × 10¹¹ m²/s² v_max = √(1.196 × 10¹¹) ≈ 3.46 × 10⁵ m/s

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 stopping potential (V₀) and the maximum kinetic energy (K_max) of the emitted photoelectrons is given by: K_max = e V₀

Given:

We can express the maximum kinetic energy in two common units: electron volts (eV) and Joules (J).

1. In electron volts (eV): The energy gained by an electron when accelerated through a potential difference of 1 Volt is 1 electron volt (1 eV). Therefore, for a stopping potential of 1.5 V, the maximum kinetic energy is directly: K_max = 1.5 eV

2. In Joules (J): Using the formula K_max = e V₀: K_max = (1.602 × 10⁻¹⁹ C) × (1.5 V) K_max = 2.403 × 10⁻¹⁹ J

Thus, the maximum kinetic energy of the photoelectrons emitted is 1.5 eV or 2.403 × 10⁻¹⁹ J.

Given:

Power (P) is the total energy emitted per second. If N is the number of photons emitted per second, each with energy E, then P = N × E. N = P / E N = (9.42 × 10⁻³ J/s) / (3.14 × 10⁻¹⁹ J/photon) N = 3 × 10¹⁶ photons/second

The momentum of the hydrogen atom (p_H) must be equal to the momentum of the photon (p). p_H = m_H × v_H Given p_H = p ≈ 1.05 × 10⁻²⁷ kg m/s

So, v_H = p / m_H v_H = (1.05 × 10⁻²⁷ kg m/s) / (1.67 × 10⁻²⁷ kg) v_H ≈ 0.63 m/s

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 the cut-off voltage (stopping potential, V₀) by: K_max = eV₀ where e is the charge of an electron.

By equating the two expressions for K_max, we get: eV₀ = hν - φ₀

To find the relationship between the cut-off voltage (V₀) and frequency (ν), we can rearrange the equation: V₀ = (h/e)ν - (φ₀/e)

This equation is in the form of a straight line, y = mx + c, where:

From the problem statement, we are given the slope of the V₀ versus ν graph. Slope (m) = 4.12 × 10⁻¹⁵ V s

We know that the slope m = h/e. Therefore, we can calculate Planck's constant (h).

Given:

h = m × e h = (4.12 × 10⁻¹⁵ V s) × (1.602 × 10⁻¹⁹ C) h ≈ 6.60 × 10⁻³⁴ J s

The calculated value of Planck's constant is approximately 6.60 × 10⁻³⁴ J s, which is very close to the accepted value.

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 is the work function of the metal.

The work function φ₀ is related to the threshold frequency ν₀ by: φ₀ = hν₀

Substituting this into the first equation, we get: K_max = hν - hν₀ = h(ν - ν₀)

Also, the maximum kinetic energy is related to the cut-off voltage V₀ by: K_max = eV₀

Combining these equations, we have: eV₀ = h(ν - ν₀)

This allows us to solve for the cut-off voltage V₀: V₀ = h(ν - ν₀) / e

Given:

Now, we substitute the values into the formula: V₀ = (6.63 × 10⁻³⁴ J s × (8.2 × 10¹⁴ Hz - 3.3 × 10¹⁴ Hz)) / (1.602 × 10⁻¹⁹ C) V₀ = (6.63 × 10⁻³⁴ × 4.9 × 10¹⁴) / (1.602 × 10⁻¹⁹) V₀ = (3.2487 × 10⁻¹⁹ J) / (1.602 × 10⁻¹⁹ C) V₀ ≈ 2.03 V

The predicted cut-off voltage for the photoelectric emission is approximately 2.03 V.

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:

First, we need to calculate the energy of the incident photons corresponding to the given wavelength. The energy of a photon (E) is given by the formula: E = hc/λ

where:

Let's calculate the energy in Joules first: E = (6.63 × 10⁻³⁴ J s × 3 × 10⁸ m/s) / (330 × 10⁻⁹ m) E = (19.89 × 10⁻²⁶ J m) / (330 × 10⁻⁹ m) E ≈ 6.027 × 10⁻¹⁹ J

Now, we need to convert this energy into electron volts (eV) to compare it with the work function. We use the conversion factor 1 eV = 1.602 × 10⁻¹⁹ J. E (in eV) = E (in J) / (1.602 × 10⁻¹⁹ J/eV) E (in eV) = (6.027 × 10⁻¹⁹ J) / (1.602 × 10⁻¹⁹ J/eV) E ≈ 3.76 eV

Now, we compare the photon energy (E) with the work function (φ₀):

Since E (3.76 eV) < φ₀ (4.2 eV), the energy of the incident photons is not sufficient to overcome the work function of the metal.

Therefore, this metal will not give photoelectric emission for incident radiation of wavelength 330 nm.

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.

Given:

Step 1: Calculate the maximum kinetic energy (K_max) The maximum kinetic energy of the ejected electrons is given by: K_max = (1/2)mv_max² K_max = (1/2) × (9.11 × 10⁻³¹ kg) × (6.0 × 10⁵ m/s)² K_max = (1/2) × (9.11 × 10⁻³¹) × (36 × 10¹⁰) K_max ≈ 1.64 × 10⁻¹⁹ J

Step 2: Calculate the energy of the incident photon (E) The energy of the incident light is given by E = hν. E = (6.63 × 10⁻³⁴ J s) × (7.21 × 10¹⁴ Hz) E ≈ 4.78 × 10⁻¹⁹ J

Step 3: Calculate the work function (φ₀) Rearranging Einstein's equation, we get: φ₀ = E - K_max φ₀ = 4.78 × 10⁻¹⁹ J - 1.64 × 10⁻¹⁹ J φ₀ = 3.14 × 10⁻¹⁹ J

Step 4: Calculate the threshold frequency (ν₀) The work function is related to the threshold frequency by the formula φ₀ = hν₀. Therefore, ν₀ = φ₀ / h. ν₀ = (3.14 × 10⁻¹⁹ J) / (6.63 × 10⁻³⁴ J s) ν₀ ≈ 4.74 × 10¹⁴ Hz

The threshold frequency for photoemission of electrons from this metal is approximately 4.74 × 10¹⁴ Hz.

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 and K_max is the maximum kinetic energy of the photoelectrons.

The energy of the incident photon (E) can be calculated from its wavelength (λ) using the formula E = hc/λ. The maximum kinetic energy (K_max) is related to the stopping potential (V₀) by K_max = eV₀.

Substituting these into the main equation, we get: hc/λ = φ₀ + eV₀

We can rearrange this equation to solve for the work function (φ₀): φ₀ = hc/λ - eV₀

Given:

Step 1: Calculate the energy of the incident photon (hc/λ) E = (6.63 × 10⁻³⁴ J s × 3 × 10⁸ m/s) / (488 × 10⁻⁹ m) E = (19.89 × 10⁻²⁶ J m) / (488 × 10⁻⁹ m) E ≈ 4.076 × 10⁻¹⁹ J

Step 2: Calculate the maximum kinetic energy (eV₀) K_max = eV₀ = (1.602 × 10⁻¹⁹ C) × (0.38 V) K_max ≈ 0.609 × 10⁻¹⁹ J

Step 3: Calculate the work function (φ₀) φ₀ = E - K_max φ₀ = 4.076 × 10⁻¹⁹ J - 0.609 × 10⁻¹⁹ J φ₀ = 3.467 × 10⁻¹⁹ J

It is common to express the work function in electron volts (eV). To do this, we divide the value in Joules by the charge of an electron: φ₀ (in eV) = (3.467 × 10⁻¹⁹ J) / (1.602 × 10⁻¹⁹ J/eV) φ₀ ≈ 2.16 eV

The work function of the material is approximately 2.16 eV.

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:

λ = h / (mv) λ = (6.63 × 10⁻³⁴ J s) / (0.040 kg × 1000 m/s) λ = (6.63 × 10⁻³⁴) / 40 λ = 1.6575 × 10⁻³⁵ m

λ = h / (mv) λ = (6.63 × 10⁻³⁴ J s) / (0.060 kg × 1.0 m/s) λ = (6.63 × 10⁻³⁴) / 0.060 λ = 1.105 × 10⁻³² m

λ = h / (mv) λ = (6.63 × 10⁻³⁴ J s) / (1.0 × 10⁻⁹ kg × 2.2 m/s) λ = (6.63 × 10⁻³⁴) / (2.2 × 10⁻⁹) λ = 3.01 × 10⁻²⁵ m

In all these cases involving macroscopic and microscopic (but not sub-atomic) objects, the de Broglie wavelength is extremely small and well beyond the limits of experimental measurement.

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.

Louis de Broglie proposed that any particle with momentum 'p' has an associated wavelength 'λ' given by: λ_deBroglie = h/p

For a photon, which is the quantum of electromagnetic radiation, its energy (E) and momentum (p) are related by: E = pc where 'c' is the speed of light. Therefore, the momentum of a photon is: p = E/c

We also know that the energy of a photon is related to the frequency (ν) of the electromagnetic radiation by Planck's relation: E = hν

Substituting this expression for E into the momentum equation: p = (hν)/c

Now, we can find the de Broglie wavelength for the photon by substituting this momentum into the de Broglie relation: λ_deBroglie = h / p = h / (hν/c) λ_deBroglie = (h * c) / (h * ν) λ_deBroglie = c/ν

For any wave, including electromagnetic radiation, the relationship between its speed (c), frequency (ν), and wavelength (λ) is given by the fundamental wave equation: c = ν * λ_wave

Solving for the wavelength, we get: λ_wave = c/ν

By comparing the results from both approaches:

We can see that λ_deBroglie = λ_wave. This shows that the wavelength of electromagnetic radiation is indeed equal to the de Broglie wavelength of its quantum (the photon). This demonstrates the consistency of the wave-particle duality concept for light.