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Class 10 Class 12

In a Rutherford's α-scattering experiment with thin gold foil, 8100 scintillations per minute are observed at an angle of 60°. What will be the number of scintillations per minute at an angle of 120°?

Given,
Number of scintillations per minute at an angle 60°, n₁ = 8100 m
Number of scintillations per minute at an angle 120° , n₂ =?

The scattering in the Rutherford's experiment is proportional to cot^{4 $\frac{ϕ}{2} .$}

\Rightarrow

\frac{n_{2}}{n_{1}} = \frac{\cot^{4} ϕ_{2} / 2}{\cot^{4} ϕ_{1} / 2}

Therefore,

^{$\frac{n_{2}}{n_{1}} = \frac{\cot^{4} (\frac{120 °}{2})}{\cot^{4} (\frac{60 °}{2})} = \frac{\cot^{4} 60 °}{\cot^{4} 30 °} = {(\frac{\frac{1}{\sqrt{3}}}{\sqrt{3}})}^{4} = \frac{1}{81}$
This implies,

$n_{2} = \frac{1}{81} \times n_{1} = \frac{1}{81} \times 8100 = 100 .$}

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A nucleus represented by the symbol _ZX^A has

Z protons and A neutrons
A protons and (Z – A) neutrons
Z neutrons and (A – Z) protons
Z protons and (A – Z) neutrons

Z protons and (A – Z) neutrons

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Can a hydrogen atom absorb a photon having energy more than 13.6 eV?

Yes, a hydrogen atom can absorb a photon having energy more than 13.6 eV. But the atom should be ionised for this to happen.

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Name the series of hydrogen spectrum lying in the infrared region.

The series of hydrogen spectrum lying in the infra-red region are Paschen series, Brackett series and Pfund series.

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An element with atomic number Z = 11 emits K-X-ray of wavelength X. The atomic number of element which emits K_a-X-ray of wavelength 4λ is

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Atoms

Can a hydrogen atom absorb a photon having energy more than 13.6 eV?