Atomic Structure - Result Question 62

####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 .

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Answer:

Correct Answer: 38. (d)

Solution:

  1. $E=\frac{h c}{\lambda}$

$\Rightarrow \frac{E _1}{E _2}=\frac{\lambda _2}{\lambda _1}=2$



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