CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Let nr and nb be respectively the number of photons emitted by a red bulb and a blue bulb of equal power in one second.


A

nr = nb

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

nr<nb

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

nr>nb

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

The information is insufficient to get a relation between nr and nb.

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

The correct option is C

nr>nb


Let νr and vb be the freqencies of red and blue light respectively. We know that vb>vr. Also, nr red photons and nb blue photons are emitted each second.

Now, since the two sources emit with equal power, the energy emitted per second will be -

nr(hvr)=nb(hvb),

which would imply,

nrvr=nbvb.

Since vb>vr, we can directly conclude: nr>nb! Option (c) is correct.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Intensity in Photon Model
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon