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A difference of 2.3 eV separates two energy levels in an atom. The frequency of radiation emitted when an atom makes transition from the upper energy level to the lower level corresponding to above mentioned energy levels is

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5.6 x 10¹⁴ Hz

6.5 x 10¹² Hz

6.5 x 10¹³ Hz

5.6 x 10¹³ Hz

answer is B.

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Detailed Solution

$ΔE = hv \Rightarrow v = \frac{2.3 \times 1.6 \times 10^{- 19}}{6.625 \times 10^{- 34}} Hz = 5.57 \times 10^{14} Hz$
$\approx$ 5.6 x 10¹⁴ Hz

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