Question

Rutherford used a 5.5 MeV alpha particle. Calculate the de Broglie wavelength of these alpha particles. How does this compare to the size of the nucleus? What if he had only been able to get alpha particles with an energy of 1 keV { would he have been able to see this e ect?

Answer 1

Wavelength of alpha particle is
`6.08 * 10^(-15)` m

The de broglie wavelength of 1keV is 451 times larger than size of nucleus.

Step-by-step explanation:

Given:

Energy of alpha particle
`E= 5.5` MeV

De broglie wavelength corresponding the energy of alpha particle is given by,

`\lambda = (h)/(√(2mE) )`

Where
`h = 6.6 * 10^(-34)` Js,
`m = 4 m_(p)`,
`m_(p) = 1.67 * 10^(-27)` kg

`\lambda = \frac{6.6 * 10^(-34) }{\sqrt{2 * 4 * 1.67 * 10^(-27) * 5.5 * 10^(6 ) * 1.6 * 10^(-19) } }`

`\lambda = 6.08 * 10^(-15)`m

Hence, wavelength of alpha particle is
`6.08 * 10^(-15)` m

Size of nucleus is ≅
`10^(-15)` m

Now wavelength of 1 keV is given by,

`\lambda _(1kev) = \frac{6.6 * 10^(-34) }{\sqrt{2 * 4 * 1.67 * 10^(-27) * 1 * 10^(3 ) * 1.6 * 10^(-19) } }`

`\lambda _(1kev) = 4.51 * 10^(-13)` m

Here,
`(\lambda_(1kev) )/(\lambda ) = (4.51 * 10^(-13) )/(10^(-15) )` ≅ 451

Therefore, the de broglie wavelength of 1keV is 451 times larger than size of nucleus.