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Q.

The wave length of 2.019 x 10¹⁶photons of light providing 1 J of energy is

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a

4000nm

b

4000 pm

c

400 pm

d

4000 m

answer is D.

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Detailed Solution

$E = \frac{nhc}{λ}$ n= number of photons

$λ = 2.019 \times 10^{16} \times 6.625 \times 10^{- 34} \times 3 \times 10^{8} = 4 \times 10^{- 9} m$

=4000 pm

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