356353
In Rutherford scattering experiment, what will be the correct angle for \(\alpha \) scattering for an impact parameter \(b = 0\)?
1 \(90^\circ \)
2 \(270^\circ \)
3 \(0^\circ \)
4 \(180^\circ \)
Explanation:
We know that impact parameter, \(b\,\alpha \,\cot \frac{\theta }{2} \Rightarrow b = K\,\cot \frac{\theta }{2}\) Here, \(K\) is constant \(b = 0\). Hence,\(\theta = 180^\circ \)
PHXII12:ATOMS
356354
In a head on collision between an \(\alpha\)-particle and gold nucleus, the closest distance of approach is \(41.3\,fm\). Calculate the energy of \(\alpha\)-particle. \((z\) of gold \(=79)\)
356355
In the Rutherford's alpha scattering experiment, as the impact parameter increases, the scattering angle of the alpha particle
1 decreases
2 increases
3 remains the same
4 is always \(90^{\circ}\)
Explanation:
Impact parameter \(b\) can be written as: \(b = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{Z{e^2}\cot (\theta /2)}}{{\left( {\frac{1}{2}m{\upsilon ^2}} \right)}}\) If impact meter increases then scattering angle decreases. Correct option is (1).
KCET - 2023
PHXII12:ATOMS
356356
In scattering experiment the force that scatters particles is
356353
In Rutherford scattering experiment, what will be the correct angle for \(\alpha \) scattering for an impact parameter \(b = 0\)?
1 \(90^\circ \)
2 \(270^\circ \)
3 \(0^\circ \)
4 \(180^\circ \)
Explanation:
We know that impact parameter, \(b\,\alpha \,\cot \frac{\theta }{2} \Rightarrow b = K\,\cot \frac{\theta }{2}\) Here, \(K\) is constant \(b = 0\). Hence,\(\theta = 180^\circ \)
PHXII12:ATOMS
356354
In a head on collision between an \(\alpha\)-particle and gold nucleus, the closest distance of approach is \(41.3\,fm\). Calculate the energy of \(\alpha\)-particle. \((z\) of gold \(=79)\)
356355
In the Rutherford's alpha scattering experiment, as the impact parameter increases, the scattering angle of the alpha particle
1 decreases
2 increases
3 remains the same
4 is always \(90^{\circ}\)
Explanation:
Impact parameter \(b\) can be written as: \(b = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{Z{e^2}\cot (\theta /2)}}{{\left( {\frac{1}{2}m{\upsilon ^2}} \right)}}\) If impact meter increases then scattering angle decreases. Correct option is (1).
KCET - 2023
PHXII12:ATOMS
356356
In scattering experiment the force that scatters particles is
356353
In Rutherford scattering experiment, what will be the correct angle for \(\alpha \) scattering for an impact parameter \(b = 0\)?
1 \(90^\circ \)
2 \(270^\circ \)
3 \(0^\circ \)
4 \(180^\circ \)
Explanation:
We know that impact parameter, \(b\,\alpha \,\cot \frac{\theta }{2} \Rightarrow b = K\,\cot \frac{\theta }{2}\) Here, \(K\) is constant \(b = 0\). Hence,\(\theta = 180^\circ \)
PHXII12:ATOMS
356354
In a head on collision between an \(\alpha\)-particle and gold nucleus, the closest distance of approach is \(41.3\,fm\). Calculate the energy of \(\alpha\)-particle. \((z\) of gold \(=79)\)
356355
In the Rutherford's alpha scattering experiment, as the impact parameter increases, the scattering angle of the alpha particle
1 decreases
2 increases
3 remains the same
4 is always \(90^{\circ}\)
Explanation:
Impact parameter \(b\) can be written as: \(b = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{Z{e^2}\cot (\theta /2)}}{{\left( {\frac{1}{2}m{\upsilon ^2}} \right)}}\) If impact meter increases then scattering angle decreases. Correct option is (1).
KCET - 2023
PHXII12:ATOMS
356356
In scattering experiment the force that scatters particles is
356353
In Rutherford scattering experiment, what will be the correct angle for \(\alpha \) scattering for an impact parameter \(b = 0\)?
1 \(90^\circ \)
2 \(270^\circ \)
3 \(0^\circ \)
4 \(180^\circ \)
Explanation:
We know that impact parameter, \(b\,\alpha \,\cot \frac{\theta }{2} \Rightarrow b = K\,\cot \frac{\theta }{2}\) Here, \(K\) is constant \(b = 0\). Hence,\(\theta = 180^\circ \)
PHXII12:ATOMS
356354
In a head on collision between an \(\alpha\)-particle and gold nucleus, the closest distance of approach is \(41.3\,fm\). Calculate the energy of \(\alpha\)-particle. \((z\) of gold \(=79)\)
356355
In the Rutherford's alpha scattering experiment, as the impact parameter increases, the scattering angle of the alpha particle
1 decreases
2 increases
3 remains the same
4 is always \(90^{\circ}\)
Explanation:
Impact parameter \(b\) can be written as: \(b = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{Z{e^2}\cot (\theta /2)}}{{\left( {\frac{1}{2}m{\upsilon ^2}} \right)}}\) If impact meter increases then scattering angle decreases. Correct option is (1).
KCET - 2023
PHXII12:ATOMS
356356
In scattering experiment the force that scatters particles is
