(a) What is de Broglie concept of matter wave ? Evaluate de Broglie wavelength of Helium that is accelerated through 300…

(a) de Broglie’s matter-wave concept: every moving material particle has a wave associated with it. The wavelength is λ = h/p, where p is the particle momentum and h = 6.626×10⁻³⁴ J s is Planck’s constant. The wave is not a mechanical wave; its amplitude is related to the probability of finding the particle. The group velocity of the matter wave equals the particle velocity, while the phase velocity is c²/v.

For a charged particle accelerated electrostatically through potential V, kinetic energy is qV = p²/(2m), so p = √(2mqV). Thus λ = h/√(2mqV).

A neutral helium atom cannot be accelerated through V. Interpreting Helium as singly ionized helium He⁺, mass m≈4m_p and charge q=e. Given m_p=m_n=1.67×10⁻²⁷ kg, m = 4×1.67×10⁻²⁷ = 6.68×10⁻²⁷ kg, q = 1.602×10⁻¹⁹ C, V = 300 V.

Then 2mqV = 2×(6.68×10⁻²⁷ kg)×(1.602×10⁻¹⁹ C)×(300 V) = 6.4208×10⁻⁴³ kg² m² s⁻², so p = √(6.4208×10⁻⁴³) = 8.013×10⁻²² kg m s⁻¹.

Hence λ = (6.626×10⁻³⁴ J s)/(8.013×10⁻²² kg m s⁻¹) = 8.27×10⁻¹³ m = 0.827 pm.

If the bare helium nucleus He²⁺ (alpha particle) is intended, q=2e, so p is √2 times larger, and λ = 5.85×10⁻¹³ m = 0.585 pm. Kinetic energy is 300 eV or 600 eV, much smaller than the helium rest energy ≈3.73 GeV, so the non-relativistic formula is valid.

(b) For an infinite one-dimensional well, the particle is confined to -a≤x≤a. Therefore the full width is L=2a. The time-independent Schrödinger equation inside the well is -(ℏ²/2m)d²ψ/dx² = Eψ, with ψ(-a)=ψ(a)=0. The allowed wave numbers are k_n = nπ/L, n=1,2,3,... so the energy levels are E_n = ℏ²k_n²/(2m) = n²h²/(8mL²).

For the transition n=4→n=2, hν = E₄-E₂ = (16-4)h²/(8mL²) = 12h²/(8mL²) = 3h²/(2mL²). Therefore L² = 3h/(2mν).

Substitute h=6.626×10⁻³⁴ J s, m=9.11×10⁻³¹ kg, ν=3.43×10¹⁴ Hz: L² = [3×6.626×10⁻³⁴]/[2×9.11×10⁻³¹×3.43×10¹⁴] = 1.9878×10⁻³³ / 6.24946×10⁻¹⁶ = 3.18075×10⁻¹⁸ m². Thus L = √(3.18075×10⁻¹⁸) = 1.783×10⁻⁹ m.

So the width of the well is L=2a=1.78×10⁻⁹ m=1.78 nm. Equivalently, a=0.892 nm.

(c) For a proton in a magnetic field B, the nuclear magnetic moment is μ = gₙ μ_N I, where μ_N = eℏ/(2m_p) = 5.0508×10⁻²⁷ J/T is the nuclear magneton and gₙ=5.585. For proton spin I=1/2, the magnetic sublevels are m_I=±1/2. The energy separation is ΔE = gₙ μ_N B. The NMR photon frequency is ν = ΔE/h = gₙ μ_N B/h. Hence B = hν/(gₙ μ_N).

Given ν=120 MHz=1.20×10⁸ Hz, hν = (6.626×10⁻³⁴ J s)(1.20×10⁸ s⁻¹) = 7.9512×10⁻²⁶ J. Also, gₙ μ_N = 5.585×5.0508×10⁻²⁷ = 2.8209×10⁻²⁶ J/T. Therefore B = 7.9512×10⁻²⁶ / 2.8209×10⁻²⁶ = 2.82 T.

Thus the required magnetic field strength is 2.82 T (neglecting chemical shift).

(d) The Landé g-factor is obtained by projecting the total magnetic moment onto the total angular momentum J. For an electron, μ_L = -μ_B L/ℏ, μ_S = -2μ_B S/ℏ, with J=L+S. The effective g-factor is g_J = 3/2 + [S(S+1)-L(L+1)]/[2J(J+1)] or equivalently g_J = 1 + [J(J+1)+S(S+1)-L(L+1)]/[2J(J+1)].

Pure orbital angular momentum: S=0, J=L. Then g_J = 3/2 + [0-L(L+1)]/[2L(L+1)] = 3/2 - 1/2 = 1. So pure orbital motion gives g=1.

Pure spin angular momentum: L=0, J=S. Then g_J = 3/2 + [S(S+1)-0]/[2S(S+1)] = 3/2 + 1/2 = 2. So pure spin gives g=2.

For the state ³P₁: 2S+1=3 gives S=1, P gives L=1, and the subscript gives J=1. Thus g_J = 3/2 + [1(1+1)-1(1+1)]/[2×1(1+1)] = 3/2 + (2-2)/4 = 3/2.

Therefore the g-factor for ³P₁ is 3/2.

(e)(i) Use the angular-momentum commutation relations, writing Jᵤ for the component along the quantization axis: [J_x,J_y]=iℏJᵤ, [J_y,Jᵤ]=iℏJ_x, [Jᵤ,J_x]=iℏJ_y.

Given J₊ = J_x + iJ_y, J₋ = J_x - iJ_y. Then [Jᵤ,J₊] = [Jᵤ,J_x+iJ_y] = [Jᵤ,J_x] + i[Jᵤ,J_y]. Now [Jᵤ,J_x]=iℏJ_y and [Jᵤ,J_y]=-iℏJ_x. Hence [Jᵤ,J₊] = iℏJ_y + i(-iℏJ_x) = iℏJ_y + ℏJ_x = ℏ(J_x+iJ_y) = ℏJ₊.

Similarly, [Jᵤ,J₋] = [Jᵤ,J_x-iJ_y] = iℏJ_y - i(-iℏJ_x) = iℏJ_y - ℏJ_x = -ℏ(J_x-iJ_y) = -ℏJ₋. Thus [Jᵤ,J₊]=+ℏJ₊, [Jᵤ,J₋]=-ℏJ₋, or compactly [Jᵤ,J_±]=±ℏJ_±.

(e)(ii) Start from the product: J₋J₊ = (J_x-iJ_y)(J_x+iJ_y) = J_x² + iJ_xJ_y - iJ_yJ_x + J_y² = J_x² + J_y² + i[J_x,J_y]. Using [J_x,J_y]=iℏJᵤ, J₋J₊ = J_x² + J_y² + i(iℏJᵤ) = J_x² + J_y² - ℏJᵤ.

Since J² = J_x² + J_y² + Jᵤ², we have J_x² + J_y² = J² - Jᵤ². Therefore J₋J₊ = J² - Jᵤ² - ℏJᵤ.

Hence the identity is proved: J₋J₊ = J² - Jᵤ² - ℏJᵤ.