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Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed of 6.55 × 106 m/s and (b) a kinetic energy of 7.89 × 10−15 J. Enter your answers in scientific notation.

User Jozzhart
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1 Answer

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Answer:

Part A:

The proton has a smaller wavelength than the electron.


\lambda_(proton) = 6.05x10^(-14)m <
\lambda_(electron) = 1.10x10^(-10)m

Part B:

The proton has a smaller wavelength than the electron.


\lambda_(proton) = 1.29x10^(-13)m <
\lambda_(electron) = 5.525x10^(-12)m

Step-by-step explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.


\lambda = (h)/(p) (1)

Where h is the Planck's constant and p is the momentum.


\lambda = (h)/(mv) (2)

Part A

Case for the electron:


\lambda = (6.624x10^(-34) J.s)/((9.11x10^(-31)Kg)(6.55x10^(6)m/s))

But
J = Kg.m^(2)/s^(2)


\lambda = (6.624x10^(-34)Kg.m^(2)/s^(2).s)/((9.11x10^(-31)Kg)(6.55x10^(6)m/s))


\lambda = 1.10x10^(-10)m

Case for the proton:


\lambda = (6.624x10^(-34)Kg.m^(2)/s^(2).s)/((1.67x10^(-27)Kg)(6.55x10^(6)m/s))


\lambda = 6.05x10^(-14)m

Hence, the proton has a smaller wavelength than the electron.

Part B

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:


KE = (1)/(2)mv^(2)


2KE = mv^(2)


v^(2) = (2KE)/(m)


v = \sqrt{(2KE)/(m)} (3)

Case for the electron:


v = \sqrt{(2(7.89x10^(-15)J))/(9.11x10^(-31)Kg)}

but
1J = kg \cdot m^(2)/s^(2)


v = \sqrt{(2(7.89x10^(-15)kg \cdot m^(2)/s^(2)))/(9.11x10^(-31)Kg)}


v = 1.316x10^(8)m/s

Then, equation 2 can be used:


\lambda = (6.624x10^(-34)Kg.m^(2)/s^(2).s)/((9.11x10^(-31)Kg)(1.316x10^(8)m/s))


\lambda = 5.525x10^(-12)m

Case for the proton :


v = \sqrt{(2(7.89x10^(-15)J))/(1.67x10^(-27)Kg)}

But
1J = kg \cdot m^(2)/s^(2)


v = \sqrt{(2(7.89x10^(-15)kg \cdot m^(2)/s^(2)))/(1.67x10^(-27)Kg)}


v = 3.07x10^(6)m/s

Then, equation 2 can be used:


\lambda = (6.624x10^(-34)Kg.m^(2)/s^(2).s)/((1.67x10^(-27)Kg)(3.07x10^(6)m/s))


\lambda = 1.29x10^(-13)m

Hence, the proton has a smaller wavelength than the electron.

User Dongryphon
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