Hertz is units for frequency. (waves per second)
wavelength = speed/frequency
if you're given the speed use that to calculate, if not then you can probably assume it's a wave of light and use the speed of light (3x10^8 m/s) to calculate.
wavelength = (3x10^8)/(1.28x10^17)
= 0.000000002 m
= 2.34 nm
<h3>
Answer:</h3>
Lead-205 (Pb-205)
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Explanation:</h3>
<u>We are given;</u>
We are supposed to identify its product after an alpha decay;
- Polonium-209 has a mass number of 209 and an atomic number of 84.
- When an element undergoes an alpha decay, the mass number decreases by 4 while the atomic number decreases by 2.
- Therefore, when Po-209 undergoes alpha decay it results to the formation of a product with a mass number of 205 and atomic number of 82.
- The product from this decay is Pb-205, because Pb-205 has a mass number of 205 and atomic number 82.
- The equation for the decay is;
²⁰⁹₈₄Po → ²⁰⁵₈₂Pb + ⁴₂He
- Note; An alpha particle is represented by a helium nucleus, ⁴₂He.
Answer:
(a) 7.11 x 10⁻³⁷ m
(b) 1.11 x 10⁻³⁵ m
Explanation:
(a) The de Broglie wavelength is given by the expression:
λ = h/p = h/mv
where h is plancks constant, p is momentum which is equal to mass times velocity.
We have all the data required to calculate the wavelength, but first we will have to convert the velocity to m/s, and the mass to kilograms to work in metric system.
v = 19.8 mi/h x ( 1609.34 m/s ) x ( 1 h / 3600 s ) = 8.85 m/s
m = 232 lb x ( 0.454 kg/ lb ) = 105.33 kg
λ = h/ mv = 6.626 x 10⁻³⁴ J·s / ( 105.33 kg x 8.85 m/s ) = 7.11 x 10⁻³⁷ m
(b) For this part we have to use the uncertainty principle associated with wave-matter:
ΔpΔx > = h/4π
mΔvΔx > = h/4π
Δx = h/ (4π m Δv )
Again to utilize this equation we will have to convert the uncertainty in velocity to m/s for unit consistency.
Δv = 0.1 mi/h x ( 1609.34 m/mi ) x ( 1 h/ 3600 s )
= 0.045 m/s
Δx = h/ (4π m Δv ) = 6.626 x 10⁻³⁴ J·s / (4π x 105.33 kg x 0.045 m/s )
= 1.11 x 10⁻³⁵ m
This calculation shows us why we should not be talking of wavelengths associatiated with everyday macroscopic objects for we are obtaining an uncertainty of 1.11 x 10⁻³⁵ m for the position of the fullback.
Answer:
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