For a hydrogen like ion $H e^{+}$ , there is a dumb-belled symmetric state A with 2 radial nodes in excited state. The final state after de-excitation was found to be ground state of H - atom.
How many spectral lines are possible there for this transition?

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Solution

The correct option is B 6
State A is dumb-belled symmetric i.e. l = 1
Radial nodes = 2 $\Rightarrow$ n - l - 1 $\Rightarrow$ n = 4
Ground state of hydrogen atom $\Rightarrow$ n = 1
Means the transition is 4 $\Rightarrow$ 1
Hence,
The total number of lines $= \frac{(n_{2} - n_{1}) (n_{2} - n_{1} + 1)}{2}$
$= \frac{(4 - 1) (4 - 1 + 1)}{2}$
$= \frac{3 \times 4}{2}$
=6 lines

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For a hydrogen like ion He+, there is a dumb-belled symmetric state A with 2 radial nodes in excited state. The final state after de-excitation was found to be ground state of H - atom. How many spectral lines are possible there for this transition?

The correct option is B 6State A is dumb-belled symmetric i.e. l = 1 Radial nodes = 2 ⇒ n - l - 1 ⇒ n = 4 Ground state of hydrogen atom ⇒ n = 1 Means the transition is 4 ⇒ 1 Hence, The total number of lines =(n2−n1)(n2−n1+1)2 =(4−1)(4−1+1)2 =3×42 =6 lines

For a hydrogen like ion $H e^{+}$ , there is a dumb-belled symmetric state A with 2 radial nodes in excited state. The final state after de-excitation was found to be ground state of H - atom.
How many spectral lines are possible there for this transition?