Question

Calculate number of photons with a wavelength of 3000pm that provide 1joule of energy

Answer 1

Explanation:

Answer

The wavelength of light is 4000pm or 4000×10

−12

m or 4000×10

−9

m.

The energy one photon is

λ

hc

=

4×10

−9

6.626×10

−34

×3×10

8

=4.97×10

−17

.

The number of photons that will provide 1 J of energy is

4.97×10

−17

1J

J=2.01×10

18

.

Answer 2

Solution:-

➝Photon energy = hn

where h = plank's constant and n is frequency

let there be N photons.. Wave length = $\frac{c}{n}$ where c is velocity of light

= $\sf{3000×10^{-12} m}$

1 joule = Nh $\frac{c}{n}$ = N = $\frac{1×n}{(h×c)}$

N=$\sf{\frac{1×3000×10^{-12}}{(3×10^{8}×6.626 ×10^{-34})}}$

$\sf{=\frac{10^{17}}{6.626}}$

$\sf{= 1.509 ×10^{16}}$

Therefore, the number of photon required is $\sf{ 1.509 ×10^{16}}$.

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