356500
Ionisation energy of a hydrogen-like ion \(A\) is greater than that of another hydrogen like ion \(B\). If \({Z_A}\) and \({Z_B}\) represent atomic numbers of \(A\) and \(B\) then
1 \({Z_B} > {Z_A}\)
2 \({Z_A} > {Z_B}\)
3 \({Z_A} = {Z_B}\)
4 \({\rm{Can't be predicted}}\)
Explanation:
Ionisation energy of a hydrogen atom is \(E = \frac{{13.6{Z^2}eV}}{{{n^2}}}\) As \({E_A} > {E_B} \Rightarrow {Z_A} > {Z_B}\) (\(n\) is same for both)
PHXII12:ATOMS
356501
The ionisation energy of 10 times ionised sodium atom is
1 \(\frac{{13.6}}{{11}}eV\)
2 \(\frac{{13.6}}{{{{(11)}^2}}}eV\)
3 \(13.6 \times {(11)^2}eV\)
4 \(13.6eV\)
Explanation:
\({E_n} = - 13.6\frac{{{Z^2}}}{{{n^2}}}\) Here, \(Z = 11\) for \(Na\)-atom, 10 electrons are removed already, so it is 10 times ionised. For the last electron to be removed, \(\therefore {E_n} = - \frac{{13.6{{(11)}^2}}}{{{{(1)}^2}}}eV \Rightarrow {E_n} = - 13.6 \times {(11)^2}eV\)
PHXII12:ATOMS
356502
Consider the spectral line resulting from the transition \(n = 2\) to \(n = 1\) in the atoms and the ions given below. The shortest wavelength is produced by
356503
The energy levels with transitions for the atom are shown. The transitions corresponding to emission of radiation of maximum and minimum wavelength are respectively
1 \(B,C\)
2 \(A,C\)
3 \(C,D\)
4 \(A,D\)
Explanation:
As, we know that the difference of energy between two level is equal to the energy of photon \(\Delta E = hv = \frac{{hc}}{\lambda } \Rightarrow \lambda = \frac{{hc}}{{\Delta E}}\) or \(\lambda \propto \frac{1}{{\Delta E}}\) The value of \(\Delta E\) is minimum for \(A\), so radiation corresponding to \(A\) have maximum wavelength and value of \(\Delta E\) is maximum for \(D\), so radiation corresponding to \(D\) have minimum wavelength
356500
Ionisation energy of a hydrogen-like ion \(A\) is greater than that of another hydrogen like ion \(B\). If \({Z_A}\) and \({Z_B}\) represent atomic numbers of \(A\) and \(B\) then
1 \({Z_B} > {Z_A}\)
2 \({Z_A} > {Z_B}\)
3 \({Z_A} = {Z_B}\)
4 \({\rm{Can't be predicted}}\)
Explanation:
Ionisation energy of a hydrogen atom is \(E = \frac{{13.6{Z^2}eV}}{{{n^2}}}\) As \({E_A} > {E_B} \Rightarrow {Z_A} > {Z_B}\) (\(n\) is same for both)
PHXII12:ATOMS
356501
The ionisation energy of 10 times ionised sodium atom is
1 \(\frac{{13.6}}{{11}}eV\)
2 \(\frac{{13.6}}{{{{(11)}^2}}}eV\)
3 \(13.6 \times {(11)^2}eV\)
4 \(13.6eV\)
Explanation:
\({E_n} = - 13.6\frac{{{Z^2}}}{{{n^2}}}\) Here, \(Z = 11\) for \(Na\)-atom, 10 electrons are removed already, so it is 10 times ionised. For the last electron to be removed, \(\therefore {E_n} = - \frac{{13.6{{(11)}^2}}}{{{{(1)}^2}}}eV \Rightarrow {E_n} = - 13.6 \times {(11)^2}eV\)
PHXII12:ATOMS
356502
Consider the spectral line resulting from the transition \(n = 2\) to \(n = 1\) in the atoms and the ions given below. The shortest wavelength is produced by
356503
The energy levels with transitions for the atom are shown. The transitions corresponding to emission of radiation of maximum and minimum wavelength are respectively
1 \(B,C\)
2 \(A,C\)
3 \(C,D\)
4 \(A,D\)
Explanation:
As, we know that the difference of energy between two level is equal to the energy of photon \(\Delta E = hv = \frac{{hc}}{\lambda } \Rightarrow \lambda = \frac{{hc}}{{\Delta E}}\) or \(\lambda \propto \frac{1}{{\Delta E}}\) The value of \(\Delta E\) is minimum for \(A\), so radiation corresponding to \(A\) have maximum wavelength and value of \(\Delta E\) is maximum for \(D\), so radiation corresponding to \(D\) have minimum wavelength
356500
Ionisation energy of a hydrogen-like ion \(A\) is greater than that of another hydrogen like ion \(B\). If \({Z_A}\) and \({Z_B}\) represent atomic numbers of \(A\) and \(B\) then
1 \({Z_B} > {Z_A}\)
2 \({Z_A} > {Z_B}\)
3 \({Z_A} = {Z_B}\)
4 \({\rm{Can't be predicted}}\)
Explanation:
Ionisation energy of a hydrogen atom is \(E = \frac{{13.6{Z^2}eV}}{{{n^2}}}\) As \({E_A} > {E_B} \Rightarrow {Z_A} > {Z_B}\) (\(n\) is same for both)
PHXII12:ATOMS
356501
The ionisation energy of 10 times ionised sodium atom is
1 \(\frac{{13.6}}{{11}}eV\)
2 \(\frac{{13.6}}{{{{(11)}^2}}}eV\)
3 \(13.6 \times {(11)^2}eV\)
4 \(13.6eV\)
Explanation:
\({E_n} = - 13.6\frac{{{Z^2}}}{{{n^2}}}\) Here, \(Z = 11\) for \(Na\)-atom, 10 electrons are removed already, so it is 10 times ionised. For the last electron to be removed, \(\therefore {E_n} = - \frac{{13.6{{(11)}^2}}}{{{{(1)}^2}}}eV \Rightarrow {E_n} = - 13.6 \times {(11)^2}eV\)
PHXII12:ATOMS
356502
Consider the spectral line resulting from the transition \(n = 2\) to \(n = 1\) in the atoms and the ions given below. The shortest wavelength is produced by
356503
The energy levels with transitions for the atom are shown. The transitions corresponding to emission of radiation of maximum and minimum wavelength are respectively
1 \(B,C\)
2 \(A,C\)
3 \(C,D\)
4 \(A,D\)
Explanation:
As, we know that the difference of energy between two level is equal to the energy of photon \(\Delta E = hv = \frac{{hc}}{\lambda } \Rightarrow \lambda = \frac{{hc}}{{\Delta E}}\) or \(\lambda \propto \frac{1}{{\Delta E}}\) The value of \(\Delta E\) is minimum for \(A\), so radiation corresponding to \(A\) have maximum wavelength and value of \(\Delta E\) is maximum for \(D\), so radiation corresponding to \(D\) have minimum wavelength
356500
Ionisation energy of a hydrogen-like ion \(A\) is greater than that of another hydrogen like ion \(B\). If \({Z_A}\) and \({Z_B}\) represent atomic numbers of \(A\) and \(B\) then
1 \({Z_B} > {Z_A}\)
2 \({Z_A} > {Z_B}\)
3 \({Z_A} = {Z_B}\)
4 \({\rm{Can't be predicted}}\)
Explanation:
Ionisation energy of a hydrogen atom is \(E = \frac{{13.6{Z^2}eV}}{{{n^2}}}\) As \({E_A} > {E_B} \Rightarrow {Z_A} > {Z_B}\) (\(n\) is same for both)
PHXII12:ATOMS
356501
The ionisation energy of 10 times ionised sodium atom is
1 \(\frac{{13.6}}{{11}}eV\)
2 \(\frac{{13.6}}{{{{(11)}^2}}}eV\)
3 \(13.6 \times {(11)^2}eV\)
4 \(13.6eV\)
Explanation:
\({E_n} = - 13.6\frac{{{Z^2}}}{{{n^2}}}\) Here, \(Z = 11\) for \(Na\)-atom, 10 electrons are removed already, so it is 10 times ionised. For the last electron to be removed, \(\therefore {E_n} = - \frac{{13.6{{(11)}^2}}}{{{{(1)}^2}}}eV \Rightarrow {E_n} = - 13.6 \times {(11)^2}eV\)
PHXII12:ATOMS
356502
Consider the spectral line resulting from the transition \(n = 2\) to \(n = 1\) in the atoms and the ions given below. The shortest wavelength is produced by
356503
The energy levels with transitions for the atom are shown. The transitions corresponding to emission of radiation of maximum and minimum wavelength are respectively
1 \(B,C\)
2 \(A,C\)
3 \(C,D\)
4 \(A,D\)
Explanation:
As, we know that the difference of energy between two level is equal to the energy of photon \(\Delta E = hv = \frac{{hc}}{\lambda } \Rightarrow \lambda = \frac{{hc}}{{\Delta E}}\) or \(\lambda \propto \frac{1}{{\Delta E}}\) The value of \(\Delta E\) is minimum for \(A\), so radiation corresponding to \(A\) have maximum wavelength and value of \(\Delta E\) is maximum for \(D\), so radiation corresponding to \(D\) have minimum wavelength
