356362
If an electron in a hydrogen atom jumps from the 3\(rd\) orbit to the 2\(nd\) orbit, it emits a photon of wavelength \((\lambda )\). When it jumps from the 4\(th\) orbit to the 3\(rd\) orbit, the corresponding wavelength of the photon will be
356363
The diagram shows the path of four \(\alpha - \) particles of the same energy being scattered by the nucleus of an atom simultaneously. Which of these are/is not physically possible
1 \(2\,{\rm{and}}\,3\)
2 \(3\,{\rm{and}}\,4\)
3 \(4\,\,{\rm{only}}\)
4 \(1\,{\rm{and}}\,4\)
Explanation:
\(\alpha - \) particles cannot be attracted by the nucleus.
PHXII12:ATOMS
356364
An alpha nucleus of energy \(\frac{1}{2}m{v^2}\) bombard a heavy nuclear target of charge \(Ze\). Then, the distance of closest approach for the alpha nucleus will be proportional to:
1 \({v^2}\)
2 \(1/m\)
3 \(1/{v^4}\)
4 \(1/Ze\)
Explanation:
For the distance of closest approach, we can write, \(\frac{1}{2}m{v^2} = \frac{{K \times (2e)(Ze)}}{r} \Rightarrow r\alpha \frac{1}{m}\) Where \(r\) is the distance of closest approach.
PHXII12:ATOMS
356365
A proton approaches another proton at rest with speed \(v_{0}\). Assume impact parameter to be zero. Their closest distance of approach is (mass of proton is \(m\) )
1 \(\dfrac{e^{2}}{4 \pi \varepsilon m v_{0}^{2}}\)
2 \(\dfrac{e^{2}}{\pi \varepsilon_{0} m v_{0}^{2}}\)
3 \(\dfrac{e^{2}}{m v_{0}^{2}}\)
4 Zero
Explanation:
Use: Law of Conservation of Linear momentum and energy. \(i.e.,\) \(m v_{0}=2 m v\) \(\dfrac{1}{2} m v_{0}^{2}-\dfrac{1}{2} m v^{2} 2=\dfrac{1}{4 \pi \in_{0}} \dfrac{e^{2}}{d}\) \(\Rightarrow \dfrac{1}{2} m v_{0}^{2}-\dfrac{m v_{0}^{2}}{4}=\dfrac{e^{2}}{4 \pi \in_{0} d}\) \(\Rightarrow \dfrac{m v_{0}^{2}}{4}=\dfrac{e^{2}}{4 \pi \in_{0} d} \Rightarrow d=\dfrac{e^{2}}{\pi \in_{0} m v_{0}^{2}}\)
356362
If an electron in a hydrogen atom jumps from the 3\(rd\) orbit to the 2\(nd\) orbit, it emits a photon of wavelength \((\lambda )\). When it jumps from the 4\(th\) orbit to the 3\(rd\) orbit, the corresponding wavelength of the photon will be
356363
The diagram shows the path of four \(\alpha - \) particles of the same energy being scattered by the nucleus of an atom simultaneously. Which of these are/is not physically possible
1 \(2\,{\rm{and}}\,3\)
2 \(3\,{\rm{and}}\,4\)
3 \(4\,\,{\rm{only}}\)
4 \(1\,{\rm{and}}\,4\)
Explanation:
\(\alpha - \) particles cannot be attracted by the nucleus.
PHXII12:ATOMS
356364
An alpha nucleus of energy \(\frac{1}{2}m{v^2}\) bombard a heavy nuclear target of charge \(Ze\). Then, the distance of closest approach for the alpha nucleus will be proportional to:
1 \({v^2}\)
2 \(1/m\)
3 \(1/{v^4}\)
4 \(1/Ze\)
Explanation:
For the distance of closest approach, we can write, \(\frac{1}{2}m{v^2} = \frac{{K \times (2e)(Ze)}}{r} \Rightarrow r\alpha \frac{1}{m}\) Where \(r\) is the distance of closest approach.
PHXII12:ATOMS
356365
A proton approaches another proton at rest with speed \(v_{0}\). Assume impact parameter to be zero. Their closest distance of approach is (mass of proton is \(m\) )
1 \(\dfrac{e^{2}}{4 \pi \varepsilon m v_{0}^{2}}\)
2 \(\dfrac{e^{2}}{\pi \varepsilon_{0} m v_{0}^{2}}\)
3 \(\dfrac{e^{2}}{m v_{0}^{2}}\)
4 Zero
Explanation:
Use: Law of Conservation of Linear momentum and energy. \(i.e.,\) \(m v_{0}=2 m v\) \(\dfrac{1}{2} m v_{0}^{2}-\dfrac{1}{2} m v^{2} 2=\dfrac{1}{4 \pi \in_{0}} \dfrac{e^{2}}{d}\) \(\Rightarrow \dfrac{1}{2} m v_{0}^{2}-\dfrac{m v_{0}^{2}}{4}=\dfrac{e^{2}}{4 \pi \in_{0} d}\) \(\Rightarrow \dfrac{m v_{0}^{2}}{4}=\dfrac{e^{2}}{4 \pi \in_{0} d} \Rightarrow d=\dfrac{e^{2}}{\pi \in_{0} m v_{0}^{2}}\)
356362
If an electron in a hydrogen atom jumps from the 3\(rd\) orbit to the 2\(nd\) orbit, it emits a photon of wavelength \((\lambda )\). When it jumps from the 4\(th\) orbit to the 3\(rd\) orbit, the corresponding wavelength of the photon will be
356363
The diagram shows the path of four \(\alpha - \) particles of the same energy being scattered by the nucleus of an atom simultaneously. Which of these are/is not physically possible
1 \(2\,{\rm{and}}\,3\)
2 \(3\,{\rm{and}}\,4\)
3 \(4\,\,{\rm{only}}\)
4 \(1\,{\rm{and}}\,4\)
Explanation:
\(\alpha - \) particles cannot be attracted by the nucleus.
PHXII12:ATOMS
356364
An alpha nucleus of energy \(\frac{1}{2}m{v^2}\) bombard a heavy nuclear target of charge \(Ze\). Then, the distance of closest approach for the alpha nucleus will be proportional to:
1 \({v^2}\)
2 \(1/m\)
3 \(1/{v^4}\)
4 \(1/Ze\)
Explanation:
For the distance of closest approach, we can write, \(\frac{1}{2}m{v^2} = \frac{{K \times (2e)(Ze)}}{r} \Rightarrow r\alpha \frac{1}{m}\) Where \(r\) is the distance of closest approach.
PHXII12:ATOMS
356365
A proton approaches another proton at rest with speed \(v_{0}\). Assume impact parameter to be zero. Their closest distance of approach is (mass of proton is \(m\) )
1 \(\dfrac{e^{2}}{4 \pi \varepsilon m v_{0}^{2}}\)
2 \(\dfrac{e^{2}}{\pi \varepsilon_{0} m v_{0}^{2}}\)
3 \(\dfrac{e^{2}}{m v_{0}^{2}}\)
4 Zero
Explanation:
Use: Law of Conservation of Linear momentum and energy. \(i.e.,\) \(m v_{0}=2 m v\) \(\dfrac{1}{2} m v_{0}^{2}-\dfrac{1}{2} m v^{2} 2=\dfrac{1}{4 \pi \in_{0}} \dfrac{e^{2}}{d}\) \(\Rightarrow \dfrac{1}{2} m v_{0}^{2}-\dfrac{m v_{0}^{2}}{4}=\dfrac{e^{2}}{4 \pi \in_{0} d}\) \(\Rightarrow \dfrac{m v_{0}^{2}}{4}=\dfrac{e^{2}}{4 \pi \in_{0} d} \Rightarrow d=\dfrac{e^{2}}{\pi \in_{0} m v_{0}^{2}}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII12:ATOMS
356362
If an electron in a hydrogen atom jumps from the 3\(rd\) orbit to the 2\(nd\) orbit, it emits a photon of wavelength \((\lambda )\). When it jumps from the 4\(th\) orbit to the 3\(rd\) orbit, the corresponding wavelength of the photon will be
356363
The diagram shows the path of four \(\alpha - \) particles of the same energy being scattered by the nucleus of an atom simultaneously. Which of these are/is not physically possible
1 \(2\,{\rm{and}}\,3\)
2 \(3\,{\rm{and}}\,4\)
3 \(4\,\,{\rm{only}}\)
4 \(1\,{\rm{and}}\,4\)
Explanation:
\(\alpha - \) particles cannot be attracted by the nucleus.
PHXII12:ATOMS
356364
An alpha nucleus of energy \(\frac{1}{2}m{v^2}\) bombard a heavy nuclear target of charge \(Ze\). Then, the distance of closest approach for the alpha nucleus will be proportional to:
1 \({v^2}\)
2 \(1/m\)
3 \(1/{v^4}\)
4 \(1/Ze\)
Explanation:
For the distance of closest approach, we can write, \(\frac{1}{2}m{v^2} = \frac{{K \times (2e)(Ze)}}{r} \Rightarrow r\alpha \frac{1}{m}\) Where \(r\) is the distance of closest approach.
PHXII12:ATOMS
356365
A proton approaches another proton at rest with speed \(v_{0}\). Assume impact parameter to be zero. Their closest distance of approach is (mass of proton is \(m\) )
1 \(\dfrac{e^{2}}{4 \pi \varepsilon m v_{0}^{2}}\)
2 \(\dfrac{e^{2}}{\pi \varepsilon_{0} m v_{0}^{2}}\)
3 \(\dfrac{e^{2}}{m v_{0}^{2}}\)
4 Zero
Explanation:
Use: Law of Conservation of Linear momentum and energy. \(i.e.,\) \(m v_{0}=2 m v\) \(\dfrac{1}{2} m v_{0}^{2}-\dfrac{1}{2} m v^{2} 2=\dfrac{1}{4 \pi \in_{0}} \dfrac{e^{2}}{d}\) \(\Rightarrow \dfrac{1}{2} m v_{0}^{2}-\dfrac{m v_{0}^{2}}{4}=\dfrac{e^{2}}{4 \pi \in_{0} d}\) \(\Rightarrow \dfrac{m v_{0}^{2}}{4}=\dfrac{e^{2}}{4 \pi \in_{0} d} \Rightarrow d=\dfrac{e^{2}}{\pi \in_{0} m v_{0}^{2}}\)
