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Tasya [4]
3 years ago
14

What is the de Broglie wavelength of an object with a mass of 2.50 kg moving at a speed of 2.70 m/s? (Useful constant: h = 6.63×

10-34 Js.)
Physics
1 answer:
xxMikexx [17]3 years ago
4 0

Answer:

9.82 × 10^{-35} Hz

Explanation:

De Broglie equation is used to determine the wavelength of a particle (e.g electron) in motion. It is given as:

λ = \frac{h}{mv}

where: λ is the required wavelength of the moving electron, h is the Planck's constant, m is the mass of the particle, v is its speed.

Given that: h = 6.63 ×10^{-34} Js, m = 2.50 kg, v = 2.70 m/s, the wavelength, λ, can be determined as follows;

λ = \frac{h}{mv}

  = \frac{6.63*10^{-34} }{2.5*2.7}

 = \frac{6.63 * 10^{-34} }{6.75}

 = 9.8222 × 10^{-35}

The wavelength of the object is 9.82 × 10^{-35} Hz.

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A 35-mm single lens reflex (SLR) digital camera is using a lens of focal length 35.0 mm to photograph a person who is 1.80 m tal
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Answer:

a) 35.44 mm

b) 17.67 mm

Explanation:

u = Object distance =  3.6 m

v = Image distance

f = Focal length = 35 mm

h_u= Object height = 1.8 m

a) Lens Equation

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{35}-\frac{1}{3600}\\\Rightarrow \frac{1}{v}=\frac{713}{25200} \\\Rightarrow v=\frac{25200}{713}=35.34\ mm

The CCD sensor is 35.34 mm from the lens

b) Magnification

m=-\frac{v}{u}\\\Rightarrow m=-\frac{35.34}{3600}

m=\frac{h_v}{h_u}\\\Rightarrow -\frac{35.34}{3600}=\frac{h_v}{1800}\\\Rightarrow h_v=-\frac{35.34}{3600}\times 1800=-17.67\ mm

The person appears 17.67 mm tall on the sensor

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