The ratio of the power of a light source to that the light source is 2. is emmitting photons per second at 600 nm. If the wavelength of the source is 300 nm, then the number of photons per second emitted by is _____ . [2025]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
The ratio of the power of a light source $S_{1}$ to that the light source $S_{2}$ is 2. $S_{1}$ is emitting $2 \times 10^{15}$ photons per second at 600 nm. If the wavelength of the source $S_{2}$ is 300 nm, then the number of photons per second emitted by $S_{2}$ is _____ $\times 10^{14}$ . [2025]

Ans.
(5)

Since power emitted by a source is given as

$P = \frac{Total energy emitted}{time}$

$= \frac{(Energy of photon) \times Number of photons (N)}{t}$

$\frac{P_{1}}{P_{2}} = \frac{(E_{1}) n_{1}}{(E_{2}) n_{2}} = \frac{(\frac{h c}{λ_{1}}) n_{1}}{(\frac{h c}{λ_{2}}) n_{2}}$

$\frac{P_{1}}{P_{2}} = (\frac{λ_{2}}{λ_{1}}) \frac{n_{1}}{n_{2}}$

Substituting the given values

$2 = (\frac{300}{600}) \times \frac{2 \times 10^{15}}{n_{2}}$

$n_{2} = \frac{1}{2} \times 10^{15} = 5 \times 10^{14} Photons / \sec$

Want Us to Email you About Special Offers & Updates?