NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII12:ATOMS
356599
In terms of Rydberg constant \(R\), the waves number of the first Balmer line is
1 \(R\)
2 \(3 R\)
3 \(\dfrac{5 R}{36}\)
4 \(\dfrac{8 R}{9}\)
Explanation:
Wave number is given by\(v=\dfrac{1}{\lambda}=R\left(\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}}\right)\) \({\rm{Given}}\quad {n_1} = 2,{n_2} = 3{\rm{ }}\) \( \Rightarrow v = R\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right)\) \(\therefore \quad v = \frac{{5R}}{{36}}\)
PHXII12:ATOMS
356600
Statement A : Hydrogen atom consists of only one electron but its emission spectrum has many lines. Statement B : Only Lyman series is found in the absorption spectrum of hydrogen atom whereas in the emission spectrum, all the series are found.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
When the atom gets appopriate energy from outside, then this electron rises to some higher energy level. Now it can return either directly to the lower energy level or come to the lowest energy level after passing through other lower energy levels hence all possible transitions take place in the source and many lines are seen in the spectrum.
PHXII12:ATOMS
356601
The ionisation energy of the electron in the hydrogen atom in its ground state is \(13.6\,eV\). The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiations corresponds to the transistion between
1 \(n = 3\,\,{\rm{to}}\,\,n = 1\)
2 \(n = 3\,\,{\rm{to}}\,\,n = 2\)
3 \(n = 4\,\,{\rm{to}}\,\,n = 3\)
4 \(n = 2\,\,{\rm{to}}\,\,n = 1\)
Explanation:
Number of spectral lines \(N = \frac{{n\left( {n - 1} \right)}}{2} = 6\) \({n^2} - n - 12 = 0 \Rightarrow n = 4\) For maximum wavelength the electron should undergo transition from \({4^{th}}\) orbit to \({3^{rd}}\) orbit
PHXII12:ATOMS
356602
The fine structure of hydrogen spectrum can be explained by
356599
In terms of Rydberg constant \(R\), the waves number of the first Balmer line is
1 \(R\)
2 \(3 R\)
3 \(\dfrac{5 R}{36}\)
4 \(\dfrac{8 R}{9}\)
Explanation:
Wave number is given by\(v=\dfrac{1}{\lambda}=R\left(\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}}\right)\) \({\rm{Given}}\quad {n_1} = 2,{n_2} = 3{\rm{ }}\) \( \Rightarrow v = R\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right)\) \(\therefore \quad v = \frac{{5R}}{{36}}\)
PHXII12:ATOMS
356600
Statement A : Hydrogen atom consists of only one electron but its emission spectrum has many lines. Statement B : Only Lyman series is found in the absorption spectrum of hydrogen atom whereas in the emission spectrum, all the series are found.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
When the atom gets appopriate energy from outside, then this electron rises to some higher energy level. Now it can return either directly to the lower energy level or come to the lowest energy level after passing through other lower energy levels hence all possible transitions take place in the source and many lines are seen in the spectrum.
PHXII12:ATOMS
356601
The ionisation energy of the electron in the hydrogen atom in its ground state is \(13.6\,eV\). The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiations corresponds to the transistion between
1 \(n = 3\,\,{\rm{to}}\,\,n = 1\)
2 \(n = 3\,\,{\rm{to}}\,\,n = 2\)
3 \(n = 4\,\,{\rm{to}}\,\,n = 3\)
4 \(n = 2\,\,{\rm{to}}\,\,n = 1\)
Explanation:
Number of spectral lines \(N = \frac{{n\left( {n - 1} \right)}}{2} = 6\) \({n^2} - n - 12 = 0 \Rightarrow n = 4\) For maximum wavelength the electron should undergo transition from \({4^{th}}\) orbit to \({3^{rd}}\) orbit
PHXII12:ATOMS
356602
The fine structure of hydrogen spectrum can be explained by
356599
In terms of Rydberg constant \(R\), the waves number of the first Balmer line is
1 \(R\)
2 \(3 R\)
3 \(\dfrac{5 R}{36}\)
4 \(\dfrac{8 R}{9}\)
Explanation:
Wave number is given by\(v=\dfrac{1}{\lambda}=R\left(\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}}\right)\) \({\rm{Given}}\quad {n_1} = 2,{n_2} = 3{\rm{ }}\) \( \Rightarrow v = R\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right)\) \(\therefore \quad v = \frac{{5R}}{{36}}\)
PHXII12:ATOMS
356600
Statement A : Hydrogen atom consists of only one electron but its emission spectrum has many lines. Statement B : Only Lyman series is found in the absorption spectrum of hydrogen atom whereas in the emission spectrum, all the series are found.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
When the atom gets appopriate energy from outside, then this electron rises to some higher energy level. Now it can return either directly to the lower energy level or come to the lowest energy level after passing through other lower energy levels hence all possible transitions take place in the source and many lines are seen in the spectrum.
PHXII12:ATOMS
356601
The ionisation energy of the electron in the hydrogen atom in its ground state is \(13.6\,eV\). The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiations corresponds to the transistion between
1 \(n = 3\,\,{\rm{to}}\,\,n = 1\)
2 \(n = 3\,\,{\rm{to}}\,\,n = 2\)
3 \(n = 4\,\,{\rm{to}}\,\,n = 3\)
4 \(n = 2\,\,{\rm{to}}\,\,n = 1\)
Explanation:
Number of spectral lines \(N = \frac{{n\left( {n - 1} \right)}}{2} = 6\) \({n^2} - n - 12 = 0 \Rightarrow n = 4\) For maximum wavelength the electron should undergo transition from \({4^{th}}\) orbit to \({3^{rd}}\) orbit
PHXII12:ATOMS
356602
The fine structure of hydrogen spectrum can be explained by
356599
In terms of Rydberg constant \(R\), the waves number of the first Balmer line is
1 \(R\)
2 \(3 R\)
3 \(\dfrac{5 R}{36}\)
4 \(\dfrac{8 R}{9}\)
Explanation:
Wave number is given by\(v=\dfrac{1}{\lambda}=R\left(\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}}\right)\) \({\rm{Given}}\quad {n_1} = 2,{n_2} = 3{\rm{ }}\) \( \Rightarrow v = R\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right)\) \(\therefore \quad v = \frac{{5R}}{{36}}\)
PHXII12:ATOMS
356600
Statement A : Hydrogen atom consists of only one electron but its emission spectrum has many lines. Statement B : Only Lyman series is found in the absorption spectrum of hydrogen atom whereas in the emission spectrum, all the series are found.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
When the atom gets appopriate energy from outside, then this electron rises to some higher energy level. Now it can return either directly to the lower energy level or come to the lowest energy level after passing through other lower energy levels hence all possible transitions take place in the source and many lines are seen in the spectrum.
PHXII12:ATOMS
356601
The ionisation energy of the electron in the hydrogen atom in its ground state is \(13.6\,eV\). The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiations corresponds to the transistion between
1 \(n = 3\,\,{\rm{to}}\,\,n = 1\)
2 \(n = 3\,\,{\rm{to}}\,\,n = 2\)
3 \(n = 4\,\,{\rm{to}}\,\,n = 3\)
4 \(n = 2\,\,{\rm{to}}\,\,n = 1\)
Explanation:
Number of spectral lines \(N = \frac{{n\left( {n - 1} \right)}}{2} = 6\) \({n^2} - n - 12 = 0 \Rightarrow n = 4\) For maximum wavelength the electron should undergo transition from \({4^{th}}\) orbit to \({3^{rd}}\) orbit
PHXII12:ATOMS
356602
The fine structure of hydrogen spectrum can be explained by
