CBSE 2022 · Region 2 · Set 1 · Q2 · 2 marks
(a)Define the terms : 'impact parameter' and 'distance of closest approach' for an $\displaystyle \alpha$-particle in Geiger-Marsden scattering experiment.(ii)What will be the value of the impact parameter for scattering angle (I) $\displaystyle \theta=0^{\circ}$ and (II) $\displaystyle \theta=180^{\circ}$ ?Photoelectric emission occurs when a surface is irradiated with the radiation of frequency (i) $\displaystyle v_{1}$, and (ii) $\displaystyle v_{2}$. The maximum kinetic energy of the electrons emitted in the two cases are K and 2K respectively. Obtain the expression for the threshold frequency for the surface.
(a)
Define the terms : 'impact parameter' and 'distance of closest approach' for an $\displaystyle \alpha$-particle in Geiger-Marsden scattering experiment.
(ii)
What will be the value of the impact parameter for scattering angle (I) $\displaystyle \theta=0^{\circ}$ and (II) $\displaystyle \theta=180^{\circ}$ ?
Photoelectric emission occurs when a surface is irradiated with the radiation of frequency (i) $\displaystyle v_{1}$, and (ii) $\displaystyle v_{2}$. The maximum kinetic energy of the electrons emitted in the two cases are K and 2K respectively. Obtain the expression for the threshold frequency for the surface.
Marking-scheme solution
(a)
Impact Parameter : It is the perpendicular distance of the initial velocity vector of the approaching $\displaystyle \alpha$-particle from the centre of the nucleus.
Distance of closest approach : It is the minimum distance of the approaching $\displaystyle \alpha$-particle and the target gold nucleus
$\displaystyle d=\frac{2 Z e^{2}}{4 \pi \varepsilon_{0} K}$ ; Where K is the kinetic energy
Alternatively : Distance of closest approach is the distance of the alpha particle from the centre of gold nucleus where its whole kinetic energy is converted into potential energy
(ii)
$\displaystyle \theta=0^{\circ}$ ; $\displaystyle b=$ maximum / almost of atomic size
$\displaystyle \theta=180^{\circ}$ ; $\displaystyle b=$ minimum $\displaystyle =$ zero
$\displaystyle K=h \nu_{1}-\phi_{0}$ and $\displaystyle 2 K=h \nu_{2}-\phi_{0}$
$\displaystyle \Rightarrow 2\left(h \nu_{1}-\phi_{0}\right)=h \nu_{2}-\phi_{0}$
$\displaystyle \Rightarrow 2 h \nu_{1}-2 \phi_{0}=h \nu_{2}-\phi_{0}$
$\displaystyle \Rightarrow h\left(2 \nu_{1}-\nu_{2}\right)=\phi_{0}=h \nu_{0}$
$\displaystyle \Rightarrow\left(2 \nu_{1}-\nu_{2}\right)=\phi_{0}=\nu_{0}$
$\displaystyle \Rightarrow \nu_{0}=2 \nu_{1}-\nu_{2}$
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CBSE Class 12 Physics past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.