####38. The ratio of the energy of a photon of $200 \AA$ wavelength radiation to that of $4000 \AA$ radiation is
$(1986,1 M)$
(a) $\frac{1}{4}$
(b) 4
(c) $\frac{1}{2}$
(d) 2 .
Correct Answer: 38. (d)
Solution:
$\Rightarrow \frac{E _1}{E _2}=\frac{\lambda _2}{\lambda _1}=2$
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