12th Class Physics Unit 18 Quantum Physics Numericals

CRQ-18.6.What should be the speed of an electron if it is to be confined in a box of the size of a nucleus (take the size of the nucleus as ~1.0 × 10⁻¹⁵ m)?

Given Δx ≈ 1.0 × 10⁻¹⁵ m, m = 9.1 × 10⁻³¹ kg, h = 6.63 × 10⁻³⁴ J s.

Using the uncertainty principle Δx·Δp ≈ h, so Δp ≈ h/Δx:
Δp = (6.63 × 10⁻³⁴ J s) / (1.0 × 10⁻¹⁵ m) = 6.63 × 10⁻¹⁹ kg m s⁻¹

Since Δp = mΔv Δv = Δp/m = (6.63 × 10⁻¹⁹ kg m s⁻¹) / (9.1 × 10⁻³¹ kg) ≈ 7.3 × 10¹¹ m s⁻¹

This required speed is far greater than the speed of light (3 × 10⁸ m s⁻¹), which is physically impossible. This confirms (as shown in the chapter for a nucleus of size 10⁻¹⁵ m) that an electron cannot be confined inside a nucleus.

Num-18.1.A photon has wavelength 350 nm. What is its energy in joules and also in eV? [Book answer: 5.68 × 10⁻¹⁹ J, 3.55 eV]

Given λ = 350 nm = 350 × 10⁻⁹ m, h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m s⁻¹

E = hc/λ = (6.63 × 10⁻³⁴ J s × 3 × 10⁸ m s⁻¹) / (350 × 10⁻⁹ m)
E = (1.989 × 10⁻²⁵ J m) / (3.5 × 10⁻⁷ m)
E ≈ 5.68 × 10⁻¹⁹ J

Converting to eV (1 eV = 1.6 × 10⁻¹⁹ J):
E = (5.68 × 10⁻¹⁹ J) / (1.6 × 10⁻¹⁹ J/eV) ≈ 3.55 eV

Num-18.2.A certain atom has a separation in energy of 3 × 10⁻¹⁸ J between two of its energy levels. What frequency photon is required to make the atom change from the lower state to the higher state? [Book answer: 4.5 × 10¹⁴ Hz]

Num-18.3.What is the wavelength of a photon with an energy of 10 × 10⁻¹⁹ J? [Book answer: 1.99 × 10⁻⁷ m]

Num-18.4.Calculate the threshold frequency for sodium metal whose work function is 2.28 eV, taking h = 6.63 × 10⁻³⁴ J s. [Book answer: 5.5 × 10¹⁴ Hz]

Given φ = 2.28 eV, h = 6.63 × 10⁻³⁴ J s, 1 eV = 1.6 × 10⁻¹⁹ J

Convert φ to joules φ = 2.28 × 1.6 × 10⁻¹⁹ J = 3.648 × 10⁻¹⁹ J

Using φ = hf₀, so f₀ = φ/h:
f₀ = (3.648 × 10⁻¹⁹ J) / (6.63 × 10⁻³⁴ J s)
f₀ ≈ 5.5 × 10¹⁴ Hz

Num-18.5.What can be the longest wavelength for photoelectric emission from tungsten if its work function is 4.54 eV? [Book answer: 273 nm]

Given φ = 4.54 eV, h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m s⁻¹, 1 eV = 1.6 × 10⁻¹⁹ J

Convert φ to joules φ = 4.54 × 1.6 × 10⁻¹⁹ J = 7.264 × 10⁻¹⁹ J

The longest wavelength (threshold wavelength λ₀) corresponds to a photon whose entire energy just equals the work function: φ = hc/λ₀, so λ₀ = hc/φ:
λ₀ = (6.63 × 10⁻³⁴ J s × 3 × 10⁸ m s⁻¹) / (7.264 × 10⁻¹⁹ J)
λ₀ ≈ 2.74 × 10⁻⁷ m ≈ 273 nm

Num-18.6.An electron is accelerated in an evacuated tube from rest through a potential difference of 550 V. What will be its final momentum? Calculate the wavelength associated with this electron. [Book answer: 1.27 × 10⁻²³ N s, 5.22 × 10⁻¹¹ m]

Given V = 550 V, e = 1.6 × 10⁻¹⁹ C, m = 9.1 × 10⁻³¹ kg, h = 6.63 × 10⁻³⁴ J s

Kinetic energy gained K.E. = eV = (1.6 × 10⁻¹⁹ C)(550 V) = 8.8 × 10⁻¹⁷ J

Since K.E. = p²/2m, momentum p = √(2m·K.E.):
p = √(2 × 9.1 × 10⁻³¹ kg × 8.8 × 10⁻¹⁷ J) = √(1.60 × 10⁻⁴⁶)
p ≈ 1.27 × 10⁻²³ kg m s⁻¹ (N s)

de Broglie wavelength: λ = h/p
λ = (6.63 × 10⁻³⁴ J s) / (1.27 × 10⁻²³ kg m s⁻¹)
λ ≈ 5.22 × 10⁻¹¹ m

Num-18.7.The position of a free electron is determined with an uncertainty of 10⁻⁷ m. What is the uncertainty in its velocity? What would be the position of the electron after one second? [Book answer: 6 × 10² m s⁻¹, anywhere within a distance of 6 × 10² m]

Given Δx = 10⁻⁷ m, h = 6.63 × 10⁻³⁴ J s, m = 9.1 × 10⁻³¹ kg

Using the (precise form of the) uncertainty principle, Δx·Δp ≥ h/4π:
Δp ≥ h/(4πΔx) = (6.63 × 10⁻³⁴ J s) / (4π × 10⁻⁷ m)
Δp ≈ 5.28 × 10⁻²⁸ kg m s⁻¹

Since Δp = mΔv Δv = Δp/m = (5.28 × 10⁻²⁸ kg m s⁻¹) / (9.1 × 10⁻³¹ kg)
Δv ≈ 5.8 × 10² m s⁻¹ ≈ 6 × 10² m s⁻¹

Position after one second the uncertainty in position grows by Δx′ = Δv × t = (6 × 10² m s⁻¹)(1 s) ≈ 6 × 10² m. So after one second, the electron could be found anywhere within a distance of about 6 × 10² m from its original position.

Num-18.8.A stationary nickel nucleus of mass 9.95 × 10⁻²⁶ kg emits a photon of energy 1.17 MeV. Find: (i) the wavelength of the photon and its momentum, (ii) the speed of the nucleus after the emission of the photon. [Book answer: (i) 1.06 × 10⁻¹² m, 6.25 × 10⁻²² N s, (ii) 6.29 × 10³ m s⁻¹]

Given E = 1.17 MeV, M = 9.95 × 10⁻²⁶ kg, h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m s⁻¹, 1 eV = 1.6 × 10⁻¹⁹ J

Convert photon energy to joules:
E = 1.17 × 10⁶ × 1.6 × 10⁻¹⁹ J = 1.872 × 10⁻¹³ J

(i) Wavelength: E = hc/λ, so λ = hc/E
λ = (6.63 × 10⁻³⁴ J s × 3 × 10⁸ m s⁻¹) / (1.872 × 10⁻¹³ J)
λ ≈ 1.06 × 10⁻¹² m

Momentum p = E/c = (1.872 × 10⁻¹³ J) / (3 × 10⁸ m s⁻¹)
p ≈ 6.24 × 10⁻²² kg m s⁻¹ (N s)

(ii) By conservation of momentum, the nucleus recoils with momentum equal in magnitude to that of the photon:
Mv = p ⟹ v = p/M = (6.24 × 10⁻²² kg m s⁻¹) / (9.95 × 10⁻²⁶ kg)
v ≈ 6.27 × 10³ m s⁻¹

Num-18.9.The work function of a metal is 3.5 eV. Light of wavelength 450 nm is incident on the surface. Find out whether electrons will be emitted by the photoelectric effect from the surface. [Book answer: Electrons will not be emitted]

Given φ = 3.5 eV, λ = 450 nm = 450 × 10⁻⁹ m, h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m s⁻¹

Energy of incident photon E = hc/λ
E = (6.63 × 10⁻³⁴ J s × 3 × 10⁸ m s⁻¹) / (450 × 10⁻⁹ m) ≈ 4.42 × 10⁻¹⁹ J

Converting to eV E = (4.42 × 10⁻¹⁹ J) / (1.6 × 10⁻¹⁹ J/eV) ≈ 2.76 eV

Since the photon energy (2.76 eV) is less than the work function (3.5 eV), each photon does not carry enough energy to free an electron from the metal surface. Therefore, no photoelectrons will be emitted, regardless of the light's intensity.

Num-18.10.A 90 keV X-ray photon is fired at a carbon target and Compton scattering occurs. Find the wavelength of the incident photon and the wavelength of the scattered photon for a scattering angle of (a) 30° and (b) 60°. [Book answer: λ = 13.8 pm; (a) 14.1 pm (b) 15 pm]

Given E = 90 keV, h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m s⁻¹, m₀(electron) = 9.1 × 10⁻³¹ kg, 1 eV = 1.6 × 10⁻¹⁹ J

Wavelength of incident photon E = 90 × 10³ × 1.6 × 10⁻¹⁹ J = 1.44 × 10⁻¹⁴ J
λ = hc/E = (6.63 × 10⁻³⁴ J s × 3 × 10⁸ m s⁻¹) / (1.44 × 10⁻¹⁴ J) ≈ 1.381 × 10⁻¹¹ m = 13.8 pm

Compton wavelength h/(m₀c) = (6.63 × 10⁻³⁴ J s) / (9.1 × 10⁻³¹ kg × 3 × 10⁸ m s⁻¹) ≈ 2.43 × 10⁻¹² m = 2.43 pm

Compton shift Δλ = (h/m₀c)(1 − cos θ)

(a) θ = 30°: 1 − cos30° = 1 − 0.866 = 0.134
Δλ = 2.43 pm × 0.134 ≈ 0.33 pm
λ′ = λ + Δλ = 13.8 pm + 0.33 pm ≈ 14.1 pm

(b) θ = 60°: 1 − cos60° = 1 − 0.5 = 0.5
Δλ = 2.43 pm × 0.5 ≈ 1.2 pm
λ′ = λ + Δλ = 13.8 pm + 1.2 pm ≈ 15.0 pm

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Unit 18: Quantum Physics — Numericals