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The de-Broglie wavelength for a He atom travelling at 1000 m/5 (typical speed at room temperature) i s __________ x10-12m•
Correct answer is '99.7'. Can you explain this answer?
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To find the de-Broglie wavelength, we can use the equation:
λ = h / p
where λ is the de-Broglie wavelength, h is the Planck's constant (6.62607015 × 10^-34 m^2 kg / s), and p is the momentum.
The momentum can be calculated using the equation:
p = m * v
where m is the mass and v is the velocity.
The mass of a helium atom (He) is approximately 4.0026 atomic mass units, which can be converted to kilograms using the conversion factor 1 atomic mass unit = 1.66054 × 10^-27 kg.
So, the mass of the helium atom (He) is:
m = 4.0026 * 1.66054 × 10^-27 kg
m ≈ 6.646175 × 10^-27 kg
The velocity is given as 1000 m/s.
Now, we can calculate the momentum:
p = m * v
p ≈ (6.646175 × 10^-27 kg) * (1000 m/s)
p ≈ 6.646175 × 10^-24 kg·m/s
Finally, we can substitute the values of h and p into the equation for the de-Broglie wavelength:
λ = h / p
λ ≈ (6.62607015 × 10^-34 m^2 kg / s) / (6.646175 × 10^-24 kg·m/s)
Simplifying the expression:
λ ≈ 9.9691 × 10^-11 m
Therefore, the de-Broglie wavelength for a He atom travelling at 1000 m/s is approximately 9.9691 × 10^-11 m or 9.9691 x 10^-12 m.
λ = h / p
where λ is the de-Broglie wavelength, h is the Planck's constant (6.62607015 × 10^-34 m^2 kg / s), and p is the momentum.
The momentum can be calculated using the equation:
p = m * v
where m is the mass and v is the velocity.
The mass of a helium atom (He) is approximately 4.0026 atomic mass units, which can be converted to kilograms using the conversion factor 1 atomic mass unit = 1.66054 × 10^-27 kg.
So, the mass of the helium atom (He) is:
m = 4.0026 * 1.66054 × 10^-27 kg
m ≈ 6.646175 × 10^-27 kg
The velocity is given as 1000 m/s.
Now, we can calculate the momentum:
p = m * v
p ≈ (6.646175 × 10^-27 kg) * (1000 m/s)
p ≈ 6.646175 × 10^-24 kg·m/s
Finally, we can substitute the values of h and p into the equation for the de-Broglie wavelength:
λ = h / p
λ ≈ (6.62607015 × 10^-34 m^2 kg / s) / (6.646175 × 10^-24 kg·m/s)
Simplifying the expression:
λ ≈ 9.9691 × 10^-11 m
Therefore, the de-Broglie wavelength for a He atom travelling at 1000 m/s is approximately 9.9691 × 10^-11 m or 9.9691 x 10^-12 m.
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The de-Broglie wavelength for a He atom travelling at 1000 m/5 (typical speed at room temperature) i s __________ x10-12m•Correct answer is '99.7'. Can you explain this answer?
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The de-Broglie wavelength for a He atom travelling at 1000 m/5 (typical speed at room temperature) i s __________ x10-12m•Correct answer is '99.7'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about The de-Broglie wavelength for a He atom travelling at 1000 m/5 (typical speed at room temperature) i s __________ x10-12m•Correct answer is '99.7'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The de-Broglie wavelength for a He atom travelling at 1000 m/5 (typical speed at room temperature) i s __________ x10-12m•Correct answer is '99.7'. Can you explain this answer?.
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