A proton, an electron, and a helium nucleus all move at speed v . Rank their de Broglie wavelengths from largest to smallest.
1 answer:
Ranking of de Broglie wavelengths from largest to smallest is electron > proton > helium
- De Broglie proposed that because light has both wave and particle properties, matter exhibits both wave and particle properties. This property has been explained as the dual behavior of matter.
- From his observations, de Broglie derived the relationship between the wavelength and momentum of matter. This relationship is known as de Broglie's relationship
De Broglie's relationship is given by .....(1) , where λ is known as de Broglie wavelength and m is mass , v is velocity , h = Plank’s constant.
From equation (1) wavelength and mass has an inverse relation .
Mass of helium is 4 times the mass of the proton and proton has a greater mass than electron.
According to equation (1) , less the mass higher will be the wavelength
Hence electron having less mass have higher wavelength and then proton and then helium having large mass will have less wavelength .
Thus, order should be electron > proton > helium .
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Answer:
P = 7500 [W]
Explanation:
First We must use Newton's second law to find the force. Newton's Second Law tells us that the sum of forces on a body is equal to the product of mass by acceleration.
∑F = m*a
where:
m = mass = 150 [kg]
a = acceleration = 5 [m/s²]
We know that the work is determined by multiplying the force by the distance, in this way the units of the work are [N*m] which corresponds to 1 Joule [J].
And the power is determined by dividing the work by the time, in this way we have that [J/s] corresponds to 1 [W].
We must convert the speed from kilometers per hour to meters per second.
where:
P = power [W]
F = force = 750 [N]
v = 10 [m/s]