You visited us 1 times! Enjoying our articles? Unlock Full Access!

Aufbau Principle

Electronic tr...

Question

Electronic transitions in H atom takes place from n₂ to n₁ shell such that:
$n_{2} + 2 n_{1} = 11$ and $2 n_{2} - 3 n_{1} = 8$
What is the total number of photons which are in Balmer series in the given transitions?

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 5
Solving $n_{2} + 2 n_{1} = 11$ and $2 n_{2} - 3 n_{1} = 8$
$n_{1} = 2$ and $n_{2} = 7$
The number of lines in the Balmer series $= 7 - 2 = 5$

Suggest Corrections

Similar questions

{H e}^{+}

n_{2}

n_{1}

2 n_{2} + 3 n_{1} = 18

2 n_{2} - 3 n_{1} = 6

n_{1}

L i^{2 +}

n_{2}

n_{1}

2 n_{2} + 3 n_{1} = 18

2 n_{2} - 3 n_{1} = 6

n_{1}

n_{1}

x + y

For which transition of $H e^{+}$ spectrum energy will be same as $2^{n d}$ line of Balmer series of H spectrum?

n = n_{1}

n = n_{2}

n_{1}

n_{2}

What electronic transition in $L i^{+ 2}$ produces the radiation of the same wave length as the first Line in the Lyman series of hydrogen?

Join BYJU'S Learning Program