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Question: A spherical copper particle has about 125 atoms. If the atomic radius of copper is 140 pm (1 pm = $1...

A spherical copper particle has about 125 atoms. If the atomic radius of copper is 140 pm (1 pm = 10−1210^{-12} m), the diameter of the particle is close to the wavelength of

A

green light

B

electron beam of energy 1.4 eV.

C

sound waves of frequency 340 Hz traveling in air at room temperature.

D

x-rays of energy 12.40 keV.

Answer

B

Explanation

Solution

To solve this problem, we need to first calculate the diameter of the copper particle and then determine the wavelength for each given option to find the closest match.

Step 1: Calculate the diameter of the spherical copper particle.

The atomic radius of copper, r=140 pm=140×10−12 mr = 140 \text{ pm} = 140 \times 10^{-12} \text{ m}. The particle is spherical and contains about 125 atoms. Assuming each atom is a sphere with radius rr, the volume of one atom is Vatom=43πr3V_{atom} = \frac{4}{3}\pi r^3. The total volume of the spherical particle, VparticleV_{particle}, containing 125 atoms, can be approximated as Vparticle=125×VatomV_{particle} = 125 \times V_{atom}. Let RR be the radius of the spherical particle. Then Vparticle=43πR3V_{particle} = \frac{4}{3}\pi R^3. Equating the two expressions for VparticleV_{particle}: 43πR3=125×43πr3\frac{4}{3}\pi R^3 = 125 \times \frac{4}{3}\pi r^3 R3=125r3R^3 = 125 r^3 Taking the cube root of both sides: R=(125)1/3rR = (125)^{1/3} r R=5rR = 5r The diameter of the particle, Dparticle=2R=2×5r=10rD_{particle} = 2R = 2 \times 5r = 10r. Substitute the value of rr: Dparticle=10×140 pm=1400 pmD_{particle} = 10 \times 140 \text{ pm} = 1400 \text{ pm} Convert pm to meters: Dparticle=1400×10−12 m=1.4×10−9 m=1.4 nmD_{particle} = 1400 \times 10^{-12} \text{ m} = 1.4 \times 10^{-9} \text{ m} = 1.4 \text{ nm}

Step 2: Calculate the wavelength for each option.

(A) Green light: The wavelength of green light typically ranges from 500 nm to 570 nm. For example, let's consider λgreen≈550 nm=5.5×10−7 m\lambda_{green} \approx 550 \text{ nm} = 5.5 \times 10^{-7} \text{ m}. This is much larger than 1.4×10−9 m1.4 \times 10^{-9} \text{ m}.

(B) Electron beam of energy 1.4 eV: The de Broglie wavelength for a particle with kinetic energy EkE_k is given by λ=h2mEk\lambda = \frac{h}{\sqrt{2mE_k}}. For an electron: Planck's constant, h=6.626×10−34 J.sh = 6.626 \times 10^{-34} \text{ J.s} Mass of electron, me=9.109×10−31 kgm_e = 9.109 \times 10^{-31} \text{ kg} Energy, Ek=1.4 eV=1.4×(1.602×10−19 J/eV)=2.2428×10−19 JE_k = 1.4 \text{ eV} = 1.4 \times (1.602 \times 10^{-19} \text{ J/eV}) = 2.2428 \times 10^{-19} \text{ J}. λelectron=6.626×10−342×9.109×10−31×2.2428×10−19\lambda_{electron} = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.109 \times 10^{-31} \times 2.2428 \times 10^{-19}}} λelectron=6.626×10−344.084×10−49=6.626×10−346.39×10−25\lambda_{electron} = \frac{6.626 \times 10^{-34}}{\sqrt{4.084 \times 10^{-49}}} = \frac{6.626 \times 10^{-34}}{6.39 \times 10^{-25}} λelectron≈1.037×10−9 m=1.037 nm\lambda_{electron} \approx 1.037 \times 10^{-9} \text{ m} = 1.037 \text{ nm} This value is close to 1.4 nm1.4 \text{ nm}.

(C) Sound waves of frequency 340 Hz traveling in air at room temperature: The speed of sound in air at room temperature, v≈340 m/sv \approx 340 \text{ m/s}. Frequency, f=340 Hzf = 340 \text{ Hz}. The wavelength is given by λ=vf\lambda = \frac{v}{f}: λsound=340 m/s340 Hz=1 m\lambda_{sound} = \frac{340 \text{ m/s}}{340 \text{ Hz}} = 1 \text{ m} This is vastly different from 1.4×10−9 m1.4 \times 10^{-9} \text{ m}.

(D) X-rays of energy 12.40 keV: For photons, the energy EE and wavelength λ\lambda are related by E=hcλE = \frac{hc}{\lambda}, so λ=hcE\lambda = \frac{hc}{E}. Planck's constant, h=6.626×10−34 J.sh = 6.626 \times 10^{-34} \text{ J.s} Speed of light, c=3×108 m/sc = 3 \times 10^8 \text{ m/s} Energy, E=12.40 keV=12.40×103 eV=12.40×103×(1.602×10−19 J/eV)=1.986×10−15 JE = 12.40 \text{ keV} = 12.40 \times 10^3 \text{ eV} = 12.40 \times 10^3 \times (1.602 \times 10^{-19} \text{ J/eV}) = 1.986 \times 10^{-15} \text{ J}. λx−ray=6.626×10−34×3×1081.986×10−15\lambda_{x-ray} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{1.986 \times 10^{-15}} λx−ray=1.9878×10−251.986×10−15≈1.00×10−10 m=0.1 nm\lambda_{x-ray} = \frac{1.9878 \times 10^{-25}}{1.986 \times 10^{-15}} \approx 1.00 \times 10^{-10} \text{ m} = 0.1 \text{ nm} This is much smaller than 1.4 nm1.4 \text{ nm}.

Step 3: Compare the particle diameter with the calculated wavelengths.

Particle diameter = 1.4 nm. (A) Green light wavelength ≈550 nm\approx 550 \text{ nm} (too large) (B) Electron beam wavelength ≈1.037 nm\approx 1.037 \text{ nm} (close) (C) Sound waves wavelength =1 m= 1 \text{ m} (too large) (D) X-rays wavelength ≈0.1 nm\approx 0.1 \text{ nm} (too small)

The diameter of the particle is closest to the wavelength of the electron beam of energy 1.4 eV.

The final answer is B\boxed{\text{B}}.

Explanation of the solution:

  1. Particle Diameter Calculation: A spherical particle with 125 atoms, each of radius rr, has a radius R=(125)1/3r=5rR = (125)^{1/3}r = 5r. Its diameter is D=2R=10rD = 2R = 10r. Given r=140 pmr = 140 \text{ pm}, D=10×140 pm=1400 pm=1.4 nmD = 10 \times 140 \text{ pm} = 1400 \text{ pm} = 1.4 \text{ nm}.
  2. Wavelength Calculations:
    • Green light: ≈550 nm\approx 550 \text{ nm}.
    • Electron beam (de Broglie): λ=h/2mE\lambda = h/\sqrt{2mE}. For E=1.4 eVE = 1.4 \text{ eV}, λ≈1.037 nm\lambda \approx 1.037 \text{ nm}.
    • Sound waves: λ=v/f\lambda = v/f. For v=340 m/sv = 340 \text{ m/s} and f=340 Hzf = 340 \text{ Hz}, λ=1 m\lambda = 1 \text{ m}.
    • X-rays: λ=hc/E\lambda = hc/E. For E=12.40 keVE = 12.40 \text{ keV}, λ≈0.1 nm\lambda \approx 0.1 \text{ nm}.
  3. Comparison: The particle diameter (1.4 nm1.4 \text{ nm}) is closest to the electron beam wavelength (1.037 nm1.037 \text{ nm}).