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Answer :
To find the wavelength of an electron with a given kinetic energy, you can use the de Broglie wavelength formula:
Wavelength (λ) = h / p
where:
λ = wavelength
h = Planck's constant (approximately 6.626 × 10⁻³⁴ joule seconds)
p = momentum of the electron = √(2mE), where m is the electron mass (approximately 9.109 × 10⁻³¹ kg) and E is the kinetic energy of the electron in joules.
Let's calculate the wavelengths for each given kinetic energy:
(a) Kinetic energy = 50.0 eV
50.0 eV = 50.0 × 1.6 × 10⁻¹⁹ J (converting eV to joules)
p = √(2 * 9.109 × 10⁻³¹ kg * 50.0 × 1.6 × 10⁻¹⁹ J)
p ≈ 3.561 × 10⁻²⁴ kg m/s
λ = 6.626 × 10⁻³⁴ J s / 3.561 × 10⁻²⁴ kg m/s ≈ 1.86 nm
(b) Kinetic energy = 500 eV
500 eV = 500 × 1.6 × 10⁻¹⁹ J
p = √(2 * 9.109 × 10⁻³¹ kg * 500 × 1.6 × 10⁻¹⁹ J)
p ≈ 1.124 × 10⁻²³ kg m/s
λ = 6.626 × 10⁻³⁴ J s / 1.124 × 10⁻²³ kg m/s ≈ 0.589 nm
(c) Kinetic energy = 5.00 keV
5.00 keV = 5.00 × 1.6 × 10⁻¹⁶ J (converting keV to joules)
p = √(2 * 9.109 × 10⁻³¹ kg * 5.00 × 1.6 × 10⁻¹⁶ J)
p ≈ 7.099 × 10⁻²² kg m/s
λ = 6.626 × 10⁻³⁴ J s / 7.099 × 10⁻²² kg m/s ≈ 0.933 nm
(d) Kinetic energy = 50.0 keV
50.0 keV = 50.0 × 1.6 × 10⁻¹⁶ J
p = √(2 * 9.109 × 10⁻³¹ kg * 50.0 × 1.6 × 10⁻¹⁶ J)
p ≈ 2.239 × 10⁻²¹ kg m/s
λ = 6.626 × 10⁻³⁴ J s / 2.239 × 10⁻²¹ kg m/s ≈ 0.296 nm
(e) Kinetic energy = 0.500 MeV
0.500 MeV = 0.500 × 1.6 × 10⁻¹³ J (converting MeV to joules)
p = √(2 * 9.109 × 10⁻³¹ kg * 0.500 × 1.6 × 10⁻¹³ J)
p ≈ 2.830 × 10⁻²⁰ kg m/s
λ = 6.626 × 10⁻³⁴ J s / 2.830 × 10⁻²⁰ kg m/s ≈ 0.234 nm
(f) Kinetic energy = 5.00 MeV
5.00 MeV = 5.00 × 1.6 × 10⁻¹³ J
p = √(2 * 9.109 × 10⁻³¹ kg * 5.00 × 1.6 × 10⁻¹³ J)
p ≈ 7.099 × 10⁻²⁰ kg m/s
λ = 6.626 × 10⁻³⁴ J s / 7.099 × 10⁻²⁰ kg m/s ≈ 0.0933 nm
Now, let's compare the wavelengths:
(a) 1.86 nm
(b) 0.589 nm
(c) 0.933 nm
(d) 0.296 nm
(e) 0.234 nm
(f) 0.0933 nm
The most suitable wavelength for studying the NaCl crystal structure is the one with the largest value because a larger wavelength allows for better diffraction and resolution in crystallography. Therefore, the most suitable wavelength is (c) 0.933 nm (5.00 keV).
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