A hydrogen atom emits $U V$ radiation of wavelength $102.5 n m$ . Find the quantum numbers of the states involved.

3, 1

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

2, 1

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

4, 1

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

4, 2

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

Open in App

Solution

The correct option is C $3, 1$
$E (e V) = \frac{1240}{102.5} = 12.1$
$E_{n} = - 13.6 + 12.1$
$= - 1.5 e V$ i.e., transition is from $n = 3$ to $n = 1$

Suggest Corrections

Similar questions

A hydrogen atom emits ultraviolet radition of wavelen gth 102.5 nm. What are the quantum numbers of the states involved in the transition ?

A hydrogen atom in a state having a binding energy of 0.85 e V makes transition to a state with excitation energy 10.2 e V.

(a) Identify the quantum numbers n of the upper and the lower energy states involved in the transition.

(b) Find the wavelength of the emitted radiation.

Join BYJU'S Learning Program