Calculate the nearest distance of approach of an α-particle of energy 2.5 eV being scattered by a gold nucleus (Z = 79).

Solution

Energy of the

α -

particle = 2.5 eV
Atomic number of the gold nucleus, Z = 79

The potential energy of an α-particle when it is at a distance x, from the nucleus is given by,

PE = (\frac{Ze}{4 {πε}_{0} x}) 2 e = \frac{2 {Ze}^{2}}{(4 {πε}_{0} x)'}

'2e' being the charge on α-particle.

Since the α-particle is momentarily stopped at a distance x, its initial kinetic energy is completely changed into potential energy here.

Hence,

\frac{1}{2} m v^{2} = \frac{2 {Ze}^{2}}{4 {πε}_{0} x} (at nearest approach K . E . = P . E .)

\Rightarrow

x = \frac{2 {Ze}^{2}}{4 {πε}_{0}} \times \frac{1}{(\frac{{mv}^{2}}{2})}

... (1)

Now energy of

α

-particle =

\frac{1}{2} m v^{2} = 2.5 MeV

= 2.5 \times 10^{6} \times 1.6 \times 10^{- 19} J = 2.5 \times 1.6 \times 10^{- 13} J

Substituting values in equation (1) we get,

x = \frac{2 \times 79 \times 1.6 \times 1.6 \times 10^{- 38} \times 9 \times 10^{9}}{2.5 \times 1.6 \times 10^{- 13}} m

= 9.101 \times 10^{- 14} m .

which is the nearest distance of approach.

Some More Questions From Wave Optics Chapter

Answer the following questions, which help you understand the difference between Thomson's model and Rutherford's model better.
Is the probability of backward scattering (i.e., scattering of α-particles at angles greater than 90°) predicted by Thomson's model much less, about the same, or much greater than that predicted by Rutherford's model?

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Wave Optics

Calculate the nearest distance of approach of an α-particle of energy 2.5 eV being scattered by a gold nucleus (Z = 79).

Some More Questions From Wave Optics Chapter

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