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 :
Solution
Given, wavelength of x ray ($\lambda$<sub>1</sub>) = 1 nm
<br><br>And wavelength of visible light ($\lambda$<sub>2</sub>) = 500 nm
<br><br>we know, Power$(P) = {{nhc} \over \lambda }$
<br><br>As P = constant and h, c also constant
<br><br>So, ${n \over \lambda } = constant$
<br><br>$\Rightarrow$ $${{{n_1}} \over {{n_2}}} = {{{\lambda _1}} \over {{\lambda _2}}} = {{1nm} \over {500nm}} = {1 \over {500}}$$
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: Photoelectric Effect
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