356603
The ratio of minimum wavelengths of Lyman and Balmer series will be
1 \(5\)
2 \(10\)
3 \(1.25\)
4 \(0.25\)
Explanation:
Wavelength is inversely proportional to energy difference and therefore minimum wavelength is for highest energy difference For Lyman series, maximum energy difference \((for\;n = 1 \to n = \infty ) = {E_0}/{1^2} = {E_0}\) For Balmer series, maximum energy difference \((for\,\,n = 2 \to n = \infty ) = {E_0}/{2^2} = {E_0}/4\) \(\therefore \) Ratio of minimum wavelengths is \(1/4 = 0.25\)
KCET - 2011
PHXII12:ATOMS
356604
A \(12.1\,eV\) electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be
1 2
2 3
3 4
4 1
Explanation:
When an electron of the beam, after hitting a hydrogen atom, completely looses its energy, the hydrogen atom jumps to the maximum excited state and its energy becomes \(( - 13 \cdot 6 + 12.1)ev\) \(i.e.\) \( - 1 \cdot 5\,ev\). \(\therefore-\dfrac{13 \cdot 6}{n^{2}}=-1 \cdot 5 \quad \Rightarrow n=3\). So the hydrogen atom jumps to the second excited state. \(\therefore\) Total number of spectral lines \(N=\dfrac{n(n-1)}{2}=\dfrac{3(3-1)}{2}=3\).
JEE - 2023
PHXII12:ATOMS
356605
In the figure, six lines of emission spectrum are shown. Which of them will be absent in the absorption spectrum?
1 \({1,2,3}\)
2 \({1,4,6}\)
3 \({4,5,6}\)
4 \({1,2,3,4,5,6}\)
Explanation:
The photon's with energies equal to that required for upward transition \({A \rightarrow X, A \rightarrow B}\) and \({A \rightarrow C}\) would be absorbed, these photons shall be absent from absorption spectrum.
PHXII12:ATOMS
356606
The energy levels of an atom is shown in figure. Which one of these transitons will result in the emission of a photon of wavelength \(124.1\,nm?\) Given \(\left( {h = 6.62 \times {{10}^{ - 34}}Js} \right)\)
356603
The ratio of minimum wavelengths of Lyman and Balmer series will be
1 \(5\)
2 \(10\)
3 \(1.25\)
4 \(0.25\)
Explanation:
Wavelength is inversely proportional to energy difference and therefore minimum wavelength is for highest energy difference For Lyman series, maximum energy difference \((for\;n = 1 \to n = \infty ) = {E_0}/{1^2} = {E_0}\) For Balmer series, maximum energy difference \((for\,\,n = 2 \to n = \infty ) = {E_0}/{2^2} = {E_0}/4\) \(\therefore \) Ratio of minimum wavelengths is \(1/4 = 0.25\)
KCET - 2011
PHXII12:ATOMS
356604
A \(12.1\,eV\) electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be
1 2
2 3
3 4
4 1
Explanation:
When an electron of the beam, after hitting a hydrogen atom, completely looses its energy, the hydrogen atom jumps to the maximum excited state and its energy becomes \(( - 13 \cdot 6 + 12.1)ev\) \(i.e.\) \( - 1 \cdot 5\,ev\). \(\therefore-\dfrac{13 \cdot 6}{n^{2}}=-1 \cdot 5 \quad \Rightarrow n=3\). So the hydrogen atom jumps to the second excited state. \(\therefore\) Total number of spectral lines \(N=\dfrac{n(n-1)}{2}=\dfrac{3(3-1)}{2}=3\).
JEE - 2023
PHXII12:ATOMS
356605
In the figure, six lines of emission spectrum are shown. Which of them will be absent in the absorption spectrum?
1 \({1,2,3}\)
2 \({1,4,6}\)
3 \({4,5,6}\)
4 \({1,2,3,4,5,6}\)
Explanation:
The photon's with energies equal to that required for upward transition \({A \rightarrow X, A \rightarrow B}\) and \({A \rightarrow C}\) would be absorbed, these photons shall be absent from absorption spectrum.
PHXII12:ATOMS
356606
The energy levels of an atom is shown in figure. Which one of these transitons will result in the emission of a photon of wavelength \(124.1\,nm?\) Given \(\left( {h = 6.62 \times {{10}^{ - 34}}Js} \right)\)
356603
The ratio of minimum wavelengths of Lyman and Balmer series will be
1 \(5\)
2 \(10\)
3 \(1.25\)
4 \(0.25\)
Explanation:
Wavelength is inversely proportional to energy difference and therefore minimum wavelength is for highest energy difference For Lyman series, maximum energy difference \((for\;n = 1 \to n = \infty ) = {E_0}/{1^2} = {E_0}\) For Balmer series, maximum energy difference \((for\,\,n = 2 \to n = \infty ) = {E_0}/{2^2} = {E_0}/4\) \(\therefore \) Ratio of minimum wavelengths is \(1/4 = 0.25\)
KCET - 2011
PHXII12:ATOMS
356604
A \(12.1\,eV\) electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be
1 2
2 3
3 4
4 1
Explanation:
When an electron of the beam, after hitting a hydrogen atom, completely looses its energy, the hydrogen atom jumps to the maximum excited state and its energy becomes \(( - 13 \cdot 6 + 12.1)ev\) \(i.e.\) \( - 1 \cdot 5\,ev\). \(\therefore-\dfrac{13 \cdot 6}{n^{2}}=-1 \cdot 5 \quad \Rightarrow n=3\). So the hydrogen atom jumps to the second excited state. \(\therefore\) Total number of spectral lines \(N=\dfrac{n(n-1)}{2}=\dfrac{3(3-1)}{2}=3\).
JEE - 2023
PHXII12:ATOMS
356605
In the figure, six lines of emission spectrum are shown. Which of them will be absent in the absorption spectrum?
1 \({1,2,3}\)
2 \({1,4,6}\)
3 \({4,5,6}\)
4 \({1,2,3,4,5,6}\)
Explanation:
The photon's with energies equal to that required for upward transition \({A \rightarrow X, A \rightarrow B}\) and \({A \rightarrow C}\) would be absorbed, these photons shall be absent from absorption spectrum.
PHXII12:ATOMS
356606
The energy levels of an atom is shown in figure. Which one of these transitons will result in the emission of a photon of wavelength \(124.1\,nm?\) Given \(\left( {h = 6.62 \times {{10}^{ - 34}}Js} \right)\)
356603
The ratio of minimum wavelengths of Lyman and Balmer series will be
1 \(5\)
2 \(10\)
3 \(1.25\)
4 \(0.25\)
Explanation:
Wavelength is inversely proportional to energy difference and therefore minimum wavelength is for highest energy difference For Lyman series, maximum energy difference \((for\;n = 1 \to n = \infty ) = {E_0}/{1^2} = {E_0}\) For Balmer series, maximum energy difference \((for\,\,n = 2 \to n = \infty ) = {E_0}/{2^2} = {E_0}/4\) \(\therefore \) Ratio of minimum wavelengths is \(1/4 = 0.25\)
KCET - 2011
PHXII12:ATOMS
356604
A \(12.1\,eV\) electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be
1 2
2 3
3 4
4 1
Explanation:
When an electron of the beam, after hitting a hydrogen atom, completely looses its energy, the hydrogen atom jumps to the maximum excited state and its energy becomes \(( - 13 \cdot 6 + 12.1)ev\) \(i.e.\) \( - 1 \cdot 5\,ev\). \(\therefore-\dfrac{13 \cdot 6}{n^{2}}=-1 \cdot 5 \quad \Rightarrow n=3\). So the hydrogen atom jumps to the second excited state. \(\therefore\) Total number of spectral lines \(N=\dfrac{n(n-1)}{2}=\dfrac{3(3-1)}{2}=3\).
JEE - 2023
PHXII12:ATOMS
356605
In the figure, six lines of emission spectrum are shown. Which of them will be absent in the absorption spectrum?
1 \({1,2,3}\)
2 \({1,4,6}\)
3 \({4,5,6}\)
4 \({1,2,3,4,5,6}\)
Explanation:
The photon's with energies equal to that required for upward transition \({A \rightarrow X, A \rightarrow B}\) and \({A \rightarrow C}\) would be absorbed, these photons shall be absent from absorption spectrum.
PHXII12:ATOMS
356606
The energy levels of an atom is shown in figure. Which one of these transitons will result in the emission of a photon of wavelength \(124.1\,nm?\) Given \(\left( {h = 6.62 \times {{10}^{ - 34}}Js} \right)\)
