In Rutherford experiments on - -ray scattering the number of particles scattered at 90- be 28 per minute- Then the number of particles scattered per minute by the same foil- but at 60- is

The correct option is A 112N ∝1/sin^4(θ/2) N_2/N_1= [ sin45/sin30 ] ^4 = [γ_2] = 2^2 N_2 = 4 × 28 = 112 ...

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Standard XII

Physics

More about Magnetic Dipoles

In Rutherford...

Question

In Rutherford experiments on $α$ -ray scattering the number of particles scattered at $90^{\circ}$ be 28 per minute. Then the number of particles scattered per minute by the same foil, but at $60^{\circ}$ is

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112

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120

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Solution

The correct option is B 112
$N \propto \frac{1}{s i n^{4} (\frac{θ}{2})}$
$\frac{N_{2}}{N_{1}} = {[\frac{s i n 45}{s i n 30}]}^{4}$
$= [γ_{2}] = 2^{2}$
$N_{2} = 4 \times 28 = 112$

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In Rutherford experiments on α-ray scattering the number of particles scattered at 90∘ be 28 per minute. Then the number of particles scattered per minute by the same foil, but at 60∘ is

The correct option is B 112N∝1sin4(θ2) N2N1=[sin45sin30]4 =[γ2]=22 N2=4×28=112

In Rutherford experiments on $α$ -ray scattering the number of particles scattered at $90^{\circ}$ be 28 per minute. Then the number of particles scattered per minute by the same foil, but at $60^{\circ}$ is

The correct option is B 112
$N \propto \frac{1}{s i n^{4} (\frac{θ}{2})}$
$\frac{N_{2}}{N_{1}} = {[\frac{s i n 45}{s i n 30}]}^{4}$
$= [γ_{2}] = 2^{2}$
$N_{2} = 4 \times 28 = 112$