Four lowest energy levels of $H$ atoms are shown in the figure The number of emission lines could be:

3

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

4

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

5

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

6

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

Open in App

Solution

The correct option is D $6$
From four lowest energy levels of H atoms, the number of emission lines could be 6.
n = 4
The number of emission lines $= \frac{n (n - 1)}{2} = \frac{4 (4 - 1)}{2} = 6$ .

Hence, the correct option is $D$

Suggest Corrections

Similar questions

Q. Four of the lowest energy levels of

H -

atom are shown in the figure. The number of possible emission lines should be,

What would be the maximum number of emission lines for atomic hydrogen that you would expect to see with the naked eye if the only electronic energy levels involved are those as shown in the Figure?

Q. In the emission line spectra of hydrogen atom, how many lines can be accounted for by all possible electron transitions between five lowest energy levels within the atom ?

Q. An atom has x energy levels; then the total number of lines in the emission spectrum will be:

Q. The figure indicates the energy level diagram of an atom and the origin of six spectral lines in emission

(

example - line number

5

arises from the transition level

B

A

)

. Which of the following spectral lines will also occur in the absorption spectrum?

Join BYJU'S Learning Program

Four lowest energy levels of H atoms are shown in the figure The number of emission lines could be:

The correct option is D 6From four lowest energy levels of H atoms, the number of emission lines could be 6. n = 4The number of emission lines =n(n−1)2=4(4−1)2=6.Hence, the correct option is D

Four lowest energy levels of $H$ atoms are shown in the figure The number of emission lines could be:

The correct option is D $6$
From four lowest energy levels of H atoms, the number of emission lines could be 6.
n = 4
The number of emission lines $= \frac{n (n - 1)}{2} = \frac{4 (4 - 1)}{2} = 6$ .

Hence, the correct option is $D$