Question
Medium
Solving time: 3 mins
In a hydrogen like atom, when an electron jumps from the -shell to the -shell, the wavelength of emitted radiation is . If an electron jumps from -shell to the -shell, the wavelength of emitted radiation will be
(a)
(b)
(c)
(d)
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For hydrogen or hydrogen like atoms, we know that
where, is Rydberg constant and is atomic number. When electron jumps from - shell to the - shell, then
for
(for } M {-shell) } \therefore \left.\frac{1}{\lambda}=R Z^{2}\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right]=\frac{5}{36} R Z^{2}\right]NLn_{1}=2 n_{2}=4 \therefore\frac{1}{\lambda^{\prime}}=R Z^{2}\left[\frac{1}{2^{2}}-\frac{1}{4^{2}}\right]=\frac{3}{16} R Z^{2}\frac{\lambda^{\prime}}{\lambda}=\left(\frac{5}{36} R Z^{2}\right) \div\left(\frac{3}{16} R Z^{2}\right)=\frac{20}{27}\lambda^{\prime}=\frac{20}{27} \lambda
where, is Rydberg constant and is atomic number. When electron jumps from - shell to the - shell, then
for
(for } M {-shell) } \therefore \left.\frac{1}{\lambda}=R Z^{2}\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right]=\frac{5}{36} R Z^{2}\right]NLn_{1}=2 n_{2}=4 \therefore\frac{1}{\lambda^{\prime}}=R Z^{2}\left[\frac{1}{2^{2}}-\frac{1}{4^{2}}\right]=\frac{3}{16} R Z^{2}\frac{\lambda^{\prime}}{\lambda}=\left(\frac{5}{36} R Z^{2}\right) \div\left(\frac{3}{16} R Z^{2}\right)=\frac{20}{27}\lambda^{\prime}=\frac{20}{27} \lambda
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| Question Text | In a hydrogen like atom, when an electron jumps from the -shell to the -shell, the wavelength of emitted radiation is . If an electron jumps from -shell to the -shell, the wavelength of emitted radiation will be (a) (b) (c) (d) |
| Updated On | Feb 28, 2023 |
| Topic | Atoms |
| Subject | Physics |
| Class | Class 12 |
| Answer Type | Text solution:1 Video solution: 2 |
| Upvotes | 234 |
| Avg. Video Duration | 8 min |
