Question

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.

Answer 1

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.