356590
If an electron in hydrogen atom jumps from an orbit of level \(n = 3\) to an orbit of level \(n = 2\), the emitted radiation has a frequency (\(R = \) Rydberg constant, \(c = \) velocity of light)
1 \(\frac{{Rc}}{{25}}\)
2 \(\frac{{5Rc}}{{36}}\)
3 \(\frac{{3Rc}}{{27}}\)
4 \(\frac{{8Rc}}{9}\)
Explanation:
When an electron jumps from higher level to lower energy level \({n_2}\), the frequency of the emitted radiation is \(v = Rc\left[ {\frac{1}{{n_2^2}} - \frac{1}{{n_1^2}}} \right]\) \(\therefore \) For \(n = 3\) to \(n = 2\) \(v = Rc\left[ {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right] = \frac{{5Rc}}{{36}}\)
KCET - 2015
PHXII12:ATOMS
356591
The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly
1 \(13.6\,eV\)
2 \(1.9\,eV\)
3 \(1.5\,eV\)
4 \(12.1\,eV\)
Explanation:
For Balmer series first transition will be from, \(n=3\) Ground state means, \(n=1\) \(\Delta E=13.6\left(\dfrac{1}{1^{2}}-\dfrac{1}{3^{2}}\right)=13.6\left(\dfrac{8}{9}\right)\) \( \Rightarrow \Delta \,E = 12.1\,eV\)
JEE - 2024
PHXII12:ATOMS
356592
The maximum number of emission lines for atomic hydrogen that you would expect to see with naked eye if the only electronic levels involved are those shown in the figure, is
1 6
2 5
3 21
4 \(\infty\)
Explanation:
Total number of emission line \(=\dfrac{n(n-1)}{2}\) \(\begin{aligned}& =\dfrac{7 \times(7-1)}{2} \\& =21 .\end{aligned}\)
PHXII12:ATOMS
356593
Hydrogen atom does not emit \(X\)-rays because:
356590
If an electron in hydrogen atom jumps from an orbit of level \(n = 3\) to an orbit of level \(n = 2\), the emitted radiation has a frequency (\(R = \) Rydberg constant, \(c = \) velocity of light)
1 \(\frac{{Rc}}{{25}}\)
2 \(\frac{{5Rc}}{{36}}\)
3 \(\frac{{3Rc}}{{27}}\)
4 \(\frac{{8Rc}}{9}\)
Explanation:
When an electron jumps from higher level to lower energy level \({n_2}\), the frequency of the emitted radiation is \(v = Rc\left[ {\frac{1}{{n_2^2}} - \frac{1}{{n_1^2}}} \right]\) \(\therefore \) For \(n = 3\) to \(n = 2\) \(v = Rc\left[ {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right] = \frac{{5Rc}}{{36}}\)
KCET - 2015
PHXII12:ATOMS
356591
The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly
1 \(13.6\,eV\)
2 \(1.9\,eV\)
3 \(1.5\,eV\)
4 \(12.1\,eV\)
Explanation:
For Balmer series first transition will be from, \(n=3\) Ground state means, \(n=1\) \(\Delta E=13.6\left(\dfrac{1}{1^{2}}-\dfrac{1}{3^{2}}\right)=13.6\left(\dfrac{8}{9}\right)\) \( \Rightarrow \Delta \,E = 12.1\,eV\)
JEE - 2024
PHXII12:ATOMS
356592
The maximum number of emission lines for atomic hydrogen that you would expect to see with naked eye if the only electronic levels involved are those shown in the figure, is
1 6
2 5
3 21
4 \(\infty\)
Explanation:
Total number of emission line \(=\dfrac{n(n-1)}{2}\) \(\begin{aligned}& =\dfrac{7 \times(7-1)}{2} \\& =21 .\end{aligned}\)
PHXII12:ATOMS
356593
Hydrogen atom does not emit \(X\)-rays because:
356590
If an electron in hydrogen atom jumps from an orbit of level \(n = 3\) to an orbit of level \(n = 2\), the emitted radiation has a frequency (\(R = \) Rydberg constant, \(c = \) velocity of light)
1 \(\frac{{Rc}}{{25}}\)
2 \(\frac{{5Rc}}{{36}}\)
3 \(\frac{{3Rc}}{{27}}\)
4 \(\frac{{8Rc}}{9}\)
Explanation:
When an electron jumps from higher level to lower energy level \({n_2}\), the frequency of the emitted radiation is \(v = Rc\left[ {\frac{1}{{n_2^2}} - \frac{1}{{n_1^2}}} \right]\) \(\therefore \) For \(n = 3\) to \(n = 2\) \(v = Rc\left[ {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right] = \frac{{5Rc}}{{36}}\)
KCET - 2015
PHXII12:ATOMS
356591
The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly
1 \(13.6\,eV\)
2 \(1.9\,eV\)
3 \(1.5\,eV\)
4 \(12.1\,eV\)
Explanation:
For Balmer series first transition will be from, \(n=3\) Ground state means, \(n=1\) \(\Delta E=13.6\left(\dfrac{1}{1^{2}}-\dfrac{1}{3^{2}}\right)=13.6\left(\dfrac{8}{9}\right)\) \( \Rightarrow \Delta \,E = 12.1\,eV\)
JEE - 2024
PHXII12:ATOMS
356592
The maximum number of emission lines for atomic hydrogen that you would expect to see with naked eye if the only electronic levels involved are those shown in the figure, is
1 6
2 5
3 21
4 \(\infty\)
Explanation:
Total number of emission line \(=\dfrac{n(n-1)}{2}\) \(\begin{aligned}& =\dfrac{7 \times(7-1)}{2} \\& =21 .\end{aligned}\)
PHXII12:ATOMS
356593
Hydrogen atom does not emit \(X\)-rays because:
356590
If an electron in hydrogen atom jumps from an orbit of level \(n = 3\) to an orbit of level \(n = 2\), the emitted radiation has a frequency (\(R = \) Rydberg constant, \(c = \) velocity of light)
1 \(\frac{{Rc}}{{25}}\)
2 \(\frac{{5Rc}}{{36}}\)
3 \(\frac{{3Rc}}{{27}}\)
4 \(\frac{{8Rc}}{9}\)
Explanation:
When an electron jumps from higher level to lower energy level \({n_2}\), the frequency of the emitted radiation is \(v = Rc\left[ {\frac{1}{{n_2^2}} - \frac{1}{{n_1^2}}} \right]\) \(\therefore \) For \(n = 3\) to \(n = 2\) \(v = Rc\left[ {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right] = \frac{{5Rc}}{{36}}\)
KCET - 2015
PHXII12:ATOMS
356591
The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly
1 \(13.6\,eV\)
2 \(1.9\,eV\)
3 \(1.5\,eV\)
4 \(12.1\,eV\)
Explanation:
For Balmer series first transition will be from, \(n=3\) Ground state means, \(n=1\) \(\Delta E=13.6\left(\dfrac{1}{1^{2}}-\dfrac{1}{3^{2}}\right)=13.6\left(\dfrac{8}{9}\right)\) \( \Rightarrow \Delta \,E = 12.1\,eV\)
JEE - 2024
PHXII12:ATOMS
356592
The maximum number of emission lines for atomic hydrogen that you would expect to see with naked eye if the only electronic levels involved are those shown in the figure, is
1 6
2 5
3 21
4 \(\infty\)
Explanation:
Total number of emission line \(=\dfrac{n(n-1)}{2}\) \(\begin{aligned}& =\dfrac{7 \times(7-1)}{2} \\& =21 .\end{aligned}\)
PHXII12:ATOMS
356593
Hydrogen atom does not emit \(X\)-rays because:
