JEE MAIN - Physics (2020 - 3rd September Evening Slot - No. 21)
Two sources of light emit X-rays of wavelength 1 nm and visible light of wavelength 500 nm, respectively. Both the sources emit light of the same power 200 W. The ratio of the number density of
photons of X-rays to the number density of photons of the visible light of the given wavelengths is :
$${1 \over {500}}$$
500
250
$${1 \over {250}}$$
Explanation
Given, wavelength of x ray ($$\lambda $$1) = 1 nm
And wavelength of visible light ($$\lambda $$2) = 500 nm
we know, Power$$(P) = {{nhc} \over \lambda }$$
As P = constant and h, c also constant
So, $${n \over \lambda } = constant$$
$$ \Rightarrow $$ $${{{n_1}} \over {{n_2}}} = {{{\lambda _1}} \over {{\lambda _2}}} = {{1nm} \over {500nm}} = {1 \over {500}}$$
And wavelength of visible light ($$\lambda $$2) = 500 nm
we know, Power$$(P) = {{nhc} \over \lambda }$$
As P = constant and h, c also constant
So, $${n \over \lambda } = constant$$
$$ \Rightarrow $$ $${{{n_1}} \over {{n_2}}} = {{{\lambda _1}} \over {{\lambda _2}}} = {{1nm} \over {500nm}} = {1 \over {500}}$$
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