NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII12:ATOMS
356451
Product of velocity and square of time period of electrons, in \({n^{th}}\) of hydrogen atom orbit is proportional to:
1 \({n^3}\)
2 \(\frac{1}{{{n^2}}}\)
3 \({n^2}\)
4 \({n^5}\)
Explanation:
We know that, the period \(\alpha \,{n^3}\) Velocity \(\alpha \,\frac{1}{n}\) \( \Rightarrow {\left( {Time\,period} \right)^2} \times velocity\,\alpha \,{n^5}\)
PHXII12:ATOMS
356452
The total energy of an electron revolving in the second orbit of hydrogen atom is
1 \( - 13.6\,eV\)
2 \( - 1.51\,eV\)
3 \( - 3.4\,eV\)
4 \({\rm{zero}}\)
Explanation:
The total energy of an electron revolving in the \({n^{th}}\) orbit of hydrogen atom is \({E_n} = - \frac{{13.6}}{{{n^2}}}eV\) For the second orbit , \(n = 2\) \(\therefore {E_2} = - \frac{{13.6}}{{{2^2}}}eV = - 3.4\,eV\)
KCET - 2018
PHXII12:ATOMS
356453
What will be the angular momentum in 4\(th\) orbit, if \(L\) is the angular momentum of the electron in the 2\(nd\) orbit of hydrogen atom?
1 \(2L\)
2 \(\frac{3}{2}L\)
3 \(\frac{2}{3}L\)
4 \(\frac{L}{2}\)
Explanation:
We know that, \({L_n} = \frac{{nh}}{{2\pi }}\) Angular momentum in 2\(nd\) orbit is \(L\). Hence,\(L = \frac{{2h}}{{2\pi }} = \frac{h}{\pi }\) The angular momentum in 4th orbit is \({L_4} = \frac{{4h}}{{2\pi }} = \frac{{2h}}{\pi } = 2L\)
PHXII12:ATOMS
356454
Magnetic field at the centre (at nucleus) of the hydrogen like atoms (atomic number \(={Z}\) ) due to the motion of electron in \(n^{\text {th }}\) orbit is proportional to \(n^{p} Z^{q}\). What is the value of \(q-p\) ?
356451
Product of velocity and square of time period of electrons, in \({n^{th}}\) of hydrogen atom orbit is proportional to:
1 \({n^3}\)
2 \(\frac{1}{{{n^2}}}\)
3 \({n^2}\)
4 \({n^5}\)
Explanation:
We know that, the period \(\alpha \,{n^3}\) Velocity \(\alpha \,\frac{1}{n}\) \( \Rightarrow {\left( {Time\,period} \right)^2} \times velocity\,\alpha \,{n^5}\)
PHXII12:ATOMS
356452
The total energy of an electron revolving in the second orbit of hydrogen atom is
1 \( - 13.6\,eV\)
2 \( - 1.51\,eV\)
3 \( - 3.4\,eV\)
4 \({\rm{zero}}\)
Explanation:
The total energy of an electron revolving in the \({n^{th}}\) orbit of hydrogen atom is \({E_n} = - \frac{{13.6}}{{{n^2}}}eV\) For the second orbit , \(n = 2\) \(\therefore {E_2} = - \frac{{13.6}}{{{2^2}}}eV = - 3.4\,eV\)
KCET - 2018
PHXII12:ATOMS
356453
What will be the angular momentum in 4\(th\) orbit, if \(L\) is the angular momentum of the electron in the 2\(nd\) orbit of hydrogen atom?
1 \(2L\)
2 \(\frac{3}{2}L\)
3 \(\frac{2}{3}L\)
4 \(\frac{L}{2}\)
Explanation:
We know that, \({L_n} = \frac{{nh}}{{2\pi }}\) Angular momentum in 2\(nd\) orbit is \(L\). Hence,\(L = \frac{{2h}}{{2\pi }} = \frac{h}{\pi }\) The angular momentum in 4th orbit is \({L_4} = \frac{{4h}}{{2\pi }} = \frac{{2h}}{\pi } = 2L\)
PHXII12:ATOMS
356454
Magnetic field at the centre (at nucleus) of the hydrogen like atoms (atomic number \(={Z}\) ) due to the motion of electron in \(n^{\text {th }}\) orbit is proportional to \(n^{p} Z^{q}\). What is the value of \(q-p\) ?
356451
Product of velocity and square of time period of electrons, in \({n^{th}}\) of hydrogen atom orbit is proportional to:
1 \({n^3}\)
2 \(\frac{1}{{{n^2}}}\)
3 \({n^2}\)
4 \({n^5}\)
Explanation:
We know that, the period \(\alpha \,{n^3}\) Velocity \(\alpha \,\frac{1}{n}\) \( \Rightarrow {\left( {Time\,period} \right)^2} \times velocity\,\alpha \,{n^5}\)
PHXII12:ATOMS
356452
The total energy of an electron revolving in the second orbit of hydrogen atom is
1 \( - 13.6\,eV\)
2 \( - 1.51\,eV\)
3 \( - 3.4\,eV\)
4 \({\rm{zero}}\)
Explanation:
The total energy of an electron revolving in the \({n^{th}}\) orbit of hydrogen atom is \({E_n} = - \frac{{13.6}}{{{n^2}}}eV\) For the second orbit , \(n = 2\) \(\therefore {E_2} = - \frac{{13.6}}{{{2^2}}}eV = - 3.4\,eV\)
KCET - 2018
PHXII12:ATOMS
356453
What will be the angular momentum in 4\(th\) orbit, if \(L\) is the angular momentum of the electron in the 2\(nd\) orbit of hydrogen atom?
1 \(2L\)
2 \(\frac{3}{2}L\)
3 \(\frac{2}{3}L\)
4 \(\frac{L}{2}\)
Explanation:
We know that, \({L_n} = \frac{{nh}}{{2\pi }}\) Angular momentum in 2\(nd\) orbit is \(L\). Hence,\(L = \frac{{2h}}{{2\pi }} = \frac{h}{\pi }\) The angular momentum in 4th orbit is \({L_4} = \frac{{4h}}{{2\pi }} = \frac{{2h}}{\pi } = 2L\)
PHXII12:ATOMS
356454
Magnetic field at the centre (at nucleus) of the hydrogen like atoms (atomic number \(={Z}\) ) due to the motion of electron in \(n^{\text {th }}\) orbit is proportional to \(n^{p} Z^{q}\). What is the value of \(q-p\) ?
356451
Product of velocity and square of time period of electrons, in \({n^{th}}\) of hydrogen atom orbit is proportional to:
1 \({n^3}\)
2 \(\frac{1}{{{n^2}}}\)
3 \({n^2}\)
4 \({n^5}\)
Explanation:
We know that, the period \(\alpha \,{n^3}\) Velocity \(\alpha \,\frac{1}{n}\) \( \Rightarrow {\left( {Time\,period} \right)^2} \times velocity\,\alpha \,{n^5}\)
PHXII12:ATOMS
356452
The total energy of an electron revolving in the second orbit of hydrogen atom is
1 \( - 13.6\,eV\)
2 \( - 1.51\,eV\)
3 \( - 3.4\,eV\)
4 \({\rm{zero}}\)
Explanation:
The total energy of an electron revolving in the \({n^{th}}\) orbit of hydrogen atom is \({E_n} = - \frac{{13.6}}{{{n^2}}}eV\) For the second orbit , \(n = 2\) \(\therefore {E_2} = - \frac{{13.6}}{{{2^2}}}eV = - 3.4\,eV\)
KCET - 2018
PHXII12:ATOMS
356453
What will be the angular momentum in 4\(th\) orbit, if \(L\) is the angular momentum of the electron in the 2\(nd\) orbit of hydrogen atom?
1 \(2L\)
2 \(\frac{3}{2}L\)
3 \(\frac{2}{3}L\)
4 \(\frac{L}{2}\)
Explanation:
We know that, \({L_n} = \frac{{nh}}{{2\pi }}\) Angular momentum in 2\(nd\) orbit is \(L\). Hence,\(L = \frac{{2h}}{{2\pi }} = \frac{h}{\pi }\) The angular momentum in 4th orbit is \({L_4} = \frac{{4h}}{{2\pi }} = \frac{{2h}}{\pi } = 2L\)
PHXII12:ATOMS
356454
Magnetic field at the centre (at nucleus) of the hydrogen like atoms (atomic number \(={Z}\) ) due to the motion of electron in \(n^{\text {th }}\) orbit is proportional to \(n^{p} Z^{q}\). What is the value of \(q-p\) ?
