KVPY SX SB Exam 20141 Question Paper

Question : 38 of 120

Marks: +1, -0

A 160 watt light source is radiating light of wavelength 6200 Å uniformly in all directions. The photon flux at a distance of 1.8 m is "of the order of (Planck’s constant 6.63 x 10

^{-34}

J-s)

$10^2\m^{-2} \s^{-1}$
$10^12\m^{-2} \s^{-1}$
$10^19\m^{-2} \s^{-1}$
$10^25\m^{-2} \s^{-1}$

Validate

Solution:

Intensity of light at 1.8 m =

P/{4π(1.8)^2

I ⇒160/{4 ×π×(1.8)^2}

Photon flux = Number of photon per unit area.
⇒

I/{{hc}/λ}

⇒

{I λ}/{{hc}}

⇒

{160× 6200 ×10^{-10}}/{4 ×π×(1.8)^2×6.63 ×10^{-34}× 3 ×10^8}

⇒

1.22 × 10^19

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